31. Height and Distance Solutions Part 4

151. The diagram is given below:

     

     We follow the question $74$ and have a similar question. Following the same method we find that $h = \sqrt{ab}. BC^2 = AC^2 + AB^2 = a^2 + ab$. Let $CD$ subtend an angle of $\alpha$ at $B$.

     In $\triangle BCD$, using sine rule

     $\frac{CD}{\sin\alpha} = \frac{BC}{\sin(90^\circ - \theta)}$

     $\Rightarrow \sin\alpha = \frac{CD}{BC}\cos\theta = \frac{CD}{BC}.\frac{AC}{BC} = \frac{(b - a)a}{a^2 + ab} = \frac{b - a}{a + b}$.

152. The diagram is given below:

     

     Let $BC$ be the pillar given a height of $h$ and $CD$ be the statue having a height of $x$. Both the statue and the pillar make the same angle at $A$ which we have let to be $\theta$.

     In $\triangle ABC, \tan\theta = \frac{h}{d}$

     In $\triangle ABD, \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{h + x}{d}$

     $\Rightarrow \frac{\frac{2h}{d}}{1 - \frac{h^2}{d^2}} = \frac{h + x}{d} \Rightarrow h + x = \frac{2hd^2}{d^2 - h^2}$

     $\Rightarrow x = \frac{h(d^2 + h^2)}{d^2 - h^2}$.

153. The diagram is given below:

     

     Let $BE$ be the tower and $CD$ be the pole such that base of the tower is at half the height of the pole. Given height of the tower is $50'$. Aangles of depression of the top and the foot of the pole from top of the tower are given as $15^\circ$ and $45^\circ$. Let the distance between the pole and the tower be $d'$.

     In $\triangle ABC, \tan45^\circ = 1 = \frac{50 + \frac{h}{2}}{d}\Rightarrow d = 50 + \frac{h}{2}$

     In $\triangle BGD, \tan15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3 + 1}} = \frac{50 - \frac{h}{2}}{d}$

     $\Rightarrow h = 100/\sqrt{3}$ ft.

154. The diagram is given below:

     

     Given $A$ is the initial point of observation and $D$ is the second point of observation which is $4$ km south of $A$. Let $P$ be the point in air where the plane is flying and $Q$ be the point directly beneath it. Given that $Q$ is directly east of $A$ and angles of elevation from $A$ and $D$ are respectively $60^\circ$ and $30^\circ$. Let $PQ = h$ km be the height of the airplane. Clearly, $\angle DAQ$ is a right angle.

     In $\triangle APQ, \tan60^\circ = \frac{h}{AQ} \Rightarrow AQ = h\cot60^\circ$

     In $\triangle DPQ, \tan30^\circ = \frac{h}{DQ} \Rightarrow DQ = h\cot30^\circ$

     In $\triangle ADQ, DQ^2 = AB^2 + AQ^2 \Rightarrow h^2\cot^230^\circ = h^2\cot^260^\circ + 4^2$

     $\Rightarrow h = \sqrt{6}$ km.

155. The diagram is given below:

     

     Let $PN$ be the flag-staff having a height of $h$. $AB$ is perpendicular to $AN$. Let $AN = x$ and $BN = y$. Given angles of elevation from $A$ and $B$ to $P$ are $\alpha$ and $\beta$ respectively.

     In $\triangle APN, x = h\cot\alpha$. In $\triangle BPN, y = h\cot\beta$

     In $\triangle ABN, AB^2 + h^2\cot^2\alpha = h^2\cot^2\beta$

     $\Rightarrow h = \frac{AB}{\sqrt{\cot^2\beta - \cot^2\alpha}} = \frac{AB\sin\alpha\sin\beta}{\sin(\alpha + \beta)\sin(\alpha - \beta)}$

156. The diagram is given below:

     

     Let $AC$ be the tower having a height of $h$ such that $AB:BC::1:9$. Given the point at a distance of $20$ m is where both $AB$ and $BC$ subtend equal angle which we have let to be $\theta$.

     In $\triangle ABD, \tan\theta = \frac{h}{10*20} \Rightarrow h = 200\tan\theta$

     In $\triangle ACD, \tan2\theta = \frac{h}{20} = 10\tan\theta$

     $\Rightarrow \frac{2\tan\theta}{1 - \tan^2\theta} = 10\tan\theta$

     $\Rightarrow 10\tan^2\theta -8 = 0 \Rightarrow \tan\theta = \frac{2}{\sqrt{5}}$

     $\Rightarrow h = 80\sqrt{5}$ m.

157. The diagram is given below:

     

     Let $BC$ be the tower inclined at angle an angle $\theta$ from horizontal having a vertical height of $h$. Let $A$ and $D$ be two equidistant points from base $B$ of the tower from where the angles of elevation to the top of the tower is $\alpha$ and $\beta$ respectively. Let $AB = BD = d$. Let $BE = x$.

     In $\triangle BCE, \tan\theta = \frac{h}{x} \Rightarrow x = h\cot\theta$

     Clearly, $AE = d - h\cot\theta$ and $BE = d + h\cot\theta$

     In $\triangle ACE, \tan\alpha = \frac{h}{d - h\cot\theta}$ and in $\triangle BCE, \tan\beta = \frac{h}{d + h\cot\theta}$

     $\Rightarrow \frac{1}{\cot\alpha + \cot\theta} = \frac{1}{\cot\beta - \cot\theta}$

     $\Rightarrow \theta = \tan^{-1}\frac{\sin(\alpha - \beta)}{2\sin\alpha\sin\beta}$.

158. The diagram is given below:

     

     Let $ABC$ be the triangle in horizontal plane and $PQ$ be the $10$ m high flag staff at the center of the $\triangle ABC$. Given that each side subtends an angle of $60^\circ$ at the top of flag staff i.e. $Q$.

     $\therefore \angle AQC = 60^\circ \Rightarrow AQ = QC$ making $\triangle AQC$ equilateral.

     Let $AQ = QC = AC = 2a$. We know that centroid is a point on median from where the top of the vertex is at a distance of $\frac{2}{3}$ rd times length of a side. We also know that median of an equilateral triangle is perpendicular bisector of the opposite side.

     $\Rightarrow AP = \frac{2}{3}.2a.\sin60^\circ = \frac{2a}{\sqrt{3}}$

     $\triangle APQ$ is also a right angle triangle with right angle at $P$.

     $\Rightarrow AQ^2 = AP^2 + PQ^2 \Rightarrow a = 5\sqrt{\frac{3}{2}}$

     $\therefore$ Length of a side $= 2a = 5\sqrt{6}$ m.

159. The diagram is given below:

     

     Let $AB$ be the pole having a height of $h$ then the height of the second pole $CD$ would be $2h$. $O$ is the point of observation situated at mid-point between the poles i.e. at a distance of $60$ m from each pole. Let $\angle AOB = \theta$ and therefore $\angle COD = 90^\circ - \theta$.

     In $\triangle AOB, \tan\theta = \frac{h}{60}$

     In $\triangle COD, \tan(90^\circ - \theta) = \cot\theta = \frac{2h}{60} = 2\tan\theta \Rightarrow \tan\theta = \frac{1}{\sqrt{2}}$

     $\Rightarrow h = 30\sqrt{2}$ m and $2h = 60\sqrt{2}$ m.

160. This problem is similar to $158$, and has been left as an exercise.

161. This problem is similar to $134$, and has been left as an exercise.

162. The diagram is given below:

     

     Since $AB$ and $CD$ are two banks of a straight river they would be parallel. We have shown alternate angles for $\beta$ and $\gamma$ in the diagram other than given angles. In $\triangle ABC, \angle ACB = \pi - (\alpha + \beta + \gamma) \Rightarrow \sin ACB = \sin(\alpha + \beta + \gamma)$.

     Using sine formula in $\triangle ABC$,

     $\frac{AB}{\sin ACB} = \frac{AC}{\sin ABC} \Rightarrow AC = \frac{a\sin\gamma}{\sin(\alpha + \beta + \gamma)}$

     Using sine formula in $\triangle ACD$,

     $\frac{CD}{\sin\alpha} = \frac{AC}{\sin\beta} \Rightarrow CD = \frac{a\sin\alpha\sin\gamma}{\sin\beta\sin(\alpha + \beta + \gamma)}$

163. The diagram is given below:

     

     Let $PQ$ be the bank of river having a width of $b$ and $R$ be the point in line with $PQ$ at a distance of $a$ from $Q$. $QS$ is the distance of $100$ m to which the person walks at right angle from initial line.

     In $\triangle PRS, \tan40^\circ = \frac{a + b}{100}$

     In $\triangle QRS, \tan25^\circ = \frac{b}{100}$

     $\Rightarrow b = 100(\tan40^\circ - \tan25^\circ)$.

164. This problem is similar to $96$, and has been left as an exercise.

165. This problem is similar to $96$, and has been left as an exercise.

166. The diagram is given below:

     

     Let $P$ and $Q$ be the tops of two spires, $P'$ and $Q'$ be their reflections. From question $OA = h$. Let $BP = BP' = h1, CQ = CQ' = h$

     Let the distance between spires be $x = MN = OM - ON$.

     In $\triangle OMP', \tan\beta = \frac{h + h1}{OM} \Rightarrow OM\tan\beta = h + h1$

     In $\triangle OMP, \tan\alpha = \frac{h1 - h}{OM} \Rightarrow OM\tan\alpha = h1 - h$

     $\Rightarrow OM(\tan\beta - \tan\alpha) = 2h \Rightarrow OM = \frac{2h}{\tan\beta - \tan\alpha}$

     Similarly, $ON = \frac{2h}{\tan\gamma - \tan\alpha}$

     $\Rightarrow x = OM - ON = 2h\left[\frac{1}{\tan\beta - \tan\alpha} - \frac{1}{\tan\gamma -\tan\alpha}\right]$

     On simplification we arrive at the desired result.

167. The diagram is given below:

     

     Let $O$ be the center of the square and $OP$ be the pole having a height of $h$. Let $OQ$ be the shdow of the pole. Given $CO = x$ and $BQ = y$. Then $BC = x + y$. Let $OR\perp BC$.

     $\therefore OR = BR = \frac{x + y}{2}$ and $QR = \frac{x - y}{2}$

     In $\triangle POQ, \tan\alpha = \frac{h}{OQ}\Rightarrow OQ = h\cot\alpha$

     In $\triangle ORQ, OQ^2 = OR^2 + QR^2 \Rightarrow h^2\cot^2\alpha = \left(\frac{x + y}{2}\right)^2 + \left(\frac{x - y}{2}\right)^2$

     $\Rightarrow h = \sqrt{\frac{x^2 + y^2}{2}}\tan\alpha$

168. The diagram is given below:

     

     Let $OP$ be the vertical height $c$ of the candle. $O'$ is the point vertically below $O$ therefore $OO' = b$ as given in the question. Let $EF$ represent the line of intersection of the wall and the horizontal ground. Draw $O'D\perp EF$ then $O'D = a$.

     Clearly, $EF = 2DE$ as shadow is symmetrical about line $O'D$,

     In similar triangles $AOP$ and $PO'E$,

     $\frac{OA}{O'E} = \frac{OP}{O'P} \Rightarrow \frac{a}{O'E} = \frac{c}{b + c} \Rightarrow O'E = \frac{a(b + c)}{c}$

     In $\triangle O'DE$,

     $O'E^2 = a^2 + DE^2 \Rightarrow DE = \frac{a}{c}\sqrt{b^2 + 2bc}$

     $\Rightarrow EF = 2DE \frac{2a}{c}\sqrt{b^2 + 2bc}$

169. The diagram is given below:

     

     Let $PABCD$ be the pyramid, $PQ$ the flag-staff having a height of $6$ m. Let $OP = h$ and the shadow touches the side at $L$.

     Proceeding like problem $167$, we have in $\triangle OML$,

     $OL^2 = OM^2 + LM^2 \Rightarrow (h + 6)^2\cot^2\alpha = \left(\frac{x + y}{2}\right)^2 + \left(\frac{x - y}{2}\right)^2$

     $\Rightarrow h = \sqrt{\frac{x^2 + y^2}{2}}\tan\alpha - 6$

170. The diagram is given below:

     

     Let $PQ$ be the tower with given height $h, C$ be the initial point of observation from where angle of elevation is $\theta$. When the man moves a distance $d$ let him reach point $B$ from where angle of elevation is $2\theta$ and then final point be $A$ which is at a distance of $\frac{3}{4}d$ from $B$, having an angle of elevation $3\theta$.

     $\angle QCB = \angle CQB = \theta \therefore BC = BQ = d$

     In $\triangle PQB, \sin2\theta = \frac{h}{d} \Rightarrow h = 2d\sin\theta\cos\theta$

     Using sine rulel in $\triangle ABQ, \frac{3d}{4\sin\theta} = \frac{d}{\sin(180^\circ - \theta)}$

     $\Rightarrow \frac{3}{4\sin\theta} = \frac{1}{3\sin\theta - 4\sin^3\theta}$

     $\Rightarrow \sin^2\theta = \frac{5}{12} \therefore \cos^2\theta = \frac{7}{12}$

     $\Rightarrow h^2 = 4d^2\sin^2\theta\cos^2\theta \Rightarrow 36h^2 = 35d^2$

171. The diagram is given below:

     

     Let $O$ be the mid-point of $AB$ having a measure of $8$ m. Let $OP$ be the $2$ m long object, $PQ$ be its position after $1$ second and $RS$ be the position after $2$ seconds.

     $\angle PAQ = \alpha, \angle RAS = \beta$ as given in the problem. Also given,

     $\frac{ds}{dt} = 2t + 1 \Rightarrow \int ds = \int(2t + 1)dt \Rightarrow s = t^2 + t + k$ where $k$ is the constant of acceleration. At $t = 0, s = 0 \Rightarrow k = 0$

     At $t = 1, s = 2$ and $t = 2, s = 6 \therefore OP = PQ = QR = RS = 2$ m.

     Let $\angle OAP = \theta_1, \angle OAQ = \theta_2, \angle OAR = \theta_3$ and $\angle OAS = \theta_4$

     $\Rightarrow \tan\alpha = \tan(\theta_2 - \theta_1) = \frac{\tan\theta_2 - \theta_1}{1 + \tan\theta_1\tan\theta_2} = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3}$

     Similarly, $\tan\beta = \frac{1}{8}$

     $\Rightarrow \cos(\alpha - \beta) = \frac{5}{\sqrt{26}}$

172. The diagram is given below:

     

     Let $OD$ be the pole having a height of $h$. Given that $\triangle ABC$ is isosceles and $B$ and $C$ subtend same angle at $P$ which is feet of the observer, therefore $AB = AC$. Let $BD = DC = x$. Given $\angle APO = \beta, \angle CPQ = \alpha$ and $OP = d$.

     $\Delta ABC = \frac{1}{2}BC.AD = x.AD$

     In $\triangle AOP, \tan\beta = \frac{h + AD}{d} \Rightarrow AD = d\tan\beta - h$

     In $\triangle CQP, \tan\alpha = \frac{h}{PQ} \Rightarrow PQ = h\cot\alpha$

     In $\triangle OPQ, OQ^2 = PQ^2 - OP^2 \Rightarrow OQ = \sqrt{h^2\cot^2\alpha - d^2}$

     $\Rightarrow \Delta ABC = (d\tan\beta - h)\sqrt{h^2\cot^2\alpha - d^2}$

173. The diagram is given below:

     

     In the diagram $A, Q, B$ are in the plane of paper and $PQ$ is perpedicular to the plane of paper.

     In $\triangle APQ, \tan(90^\circ - \theta) = \frac{h}{AQ}$

     In $\triangle BPQ, \tan\theta = \frac{h}{BQ}$

     $\Rightarrow h = \sqrt{AQ.BQ}$

     Since $BQ$ is north-west $\therefore \angle AQB = 45^\circ = \angle QBA\Rightarrow AQ = AB = 100$ m.

     In $\triangle ABQ, OB = \sqrt{AQ^2 + AB^2} = 100\sqrt{2}$ m.

     $\Rightarrow h = 100\sqrt[4]{2}$ m.

174. The diagram is given below:

     

     Let $AB$ and $CD$ be the vertical poles having heights of $a$ and $b$ respectively and angle of elevation $\alpha$ from $O$ which is same for both of them. Also, the angles of elevation from $P$ are $\beta$ and $\gamma$ along with $\angle APC = 90^\circ$.

     In $\triangle ABQ, \tan\alpha = \frac{a}{AQ} \Rightarrow AQ = a\cot\alpha$

     In $\triangle CDQ, \tan\alpha = \frac{b}{CQ} \Rightarrow CQ = b\cot\alpha$

     $\Rightarrow AC = AQ + CQ = (a + b)\cot\alpha$

     In $\triangle ABP, \tan\beta = \frac{a}{AP} \Rightarrow AP = a\cot\beta$

     In $\triangle CDP, \tan\gamma = \frac{a}{CP}\Rightarrow CP = b\cot\gamma$

     In $\triangle APC, AC^2 = AP^2 + CP^2$

     $\Rightarrow (a + b)^2\cot^2\alpha = a^2\cot^2\beta + b^2\cot^2\gamma$

175. The diagram is given below:

     

     Given the pole is $PQ$, let $h$ be its height. $PQ$ is perpedicular to the plane of paper i.e $ABC. \therefore \angle QPA = \angle QPB = \angle QPC = 90^\circ$

     In $\triangle APQ, \tan\theta = \frac{h}{PA} \Rightarrow PA = h\cot\theta$

     Similarly, $PB = PC = h\cot\theta = PA$

     Hence, $P$ is the circumcenter of the $\triangle ABC$ and $PA$ is circum-radius of the circumcircle.

     $\therefore h = PA\tan\theta = \frac{abc}{4\Delta}\tan\theta$

176. The diagram is given below:

     

     Let $PQ$ be the tower having a height of $h$ and $\angle AOP = \theta$. Given that $\tan\theta = \frac{1}{\sqrt{2}}$

     In $\triangle AOP, \tan\theta = \frac{AP}{AO} \Rightarrow AP = 150\sqrt{2}$ m.

     In $\triangle POQ, \tan30^\circ = \frac{h}{OP} \Rightarrow OP = h\sqrt{3}$

     In $\triangle AOP, OP^2 = OA^2 + AP^2 \Rightarrow 3h^2 = 300^2 + (150\sqrt{2})^2$

     $\Rightarrow h = 150\sqrt{2}$ m,

     $\therefore \tan\phi = \frac{h}{AP} = 1 \Rightarrow \phi = 45^\circ$.

177. The diagram is given below:

     

     Let $OB = h, OA = x$. In $\triangle AOB, \tan\alpha =\frac{x}{h}$

     $\Rightarrow x = h\tan\alpha$

     In $\triangle BOC, \tan\beta = \frac{h}{d - x} \Rightarrow d = h\tan\alpha + h\cot\beta$

     In $\triangle BOD, \tan\gamma = \frac{h}{d + x}\Rightarrow d = h\cot\gamma - h\tan\alpha$

     $\therefore h\tan\alpha + h\cot\beta = h\cot\gamma - h\tan\alpha$

     $\Rightarrow 2\tan\alpha = \cot\gamma - \cot\beta$.

178. The diagram is given below:

     

     Let $M$ be the mid-point of $ES$ such that $SM = ME = x$ and $OP$ be the tower having a height of $h$.

     In $\triangle EOP, \tan\alpha = \frac{h}{OE} \Rightarrow OE = h\cot\alpha$

     Similarly in $\triangle OPS, OS = h\cot\beta$ and in $\triangle MOP, OM = h\cot\theta$

     Since $OE$ is eastward and $OS$ is southward $\Rightarrow EOS = 90^\circ$

     $\Rightarrow ES^2 = OS^2 + OE^2 \Rightarrow 4x^2 = h^2(\cot^2\beta + cot^2\alpha)$

     Since $M$ is mid-point of $ES, OM$ would be the median.

     $\Rightarrow OS^2 + OE^2 = 2MS^2 + 2OM^2$

     $\Rightarrow h^2\cot^2\beta + h^2\cot^2\alpha = \frac{h^2(\cot^2\beta + \cot^2\alpha)}{2} + 2h^2\cot^2\theta$

     $\Rightarrow \cot^2\beta + \cot^2\alpha = 4\cot^2\theta$

179. The diagram is given below:

     

     Let $AP$ be the tree having a height of $h$ and $AB$ be the width of canal equal to $x$. Given, $BC = 20$ m and $\angle BAC = 120^\circ$.

     In $\triangle ABP, \tan60^\circ = \frac{h}{AB} \Rightarrow AB = \frac{h}{\sqrt{3}}$

     In $\triangle ACP, \tan30^\circ = \frac{h}{AC}\Rightarrow AC = \sqrt{3}h$

     Using cosine rule in $\triangle ABC,$

     $\cos120^\circ = \frac{AB^2 + BC^2 - AC^2}{2.AB.BC}\Rightarrow -\frac{1}{2} = \frac{\frac{h^2}{3} + 20^2 - 3h^2}{2.20.\frac{h}{\sqrt{3}}}$

     $\Rightarrow 2h^2 - 5\sqrt{3}h - 300 = 0\Rightarrow h = \frac{5\sqrt{3} + 15\sqrt{11}}{4}$ m.

     $\Rightarrow AB = \frac{5 + 5\sqrt{33}}{4}$ m.

180. The diagram is given below:

     

     Let $OP$ be the tower with $P$ being the top having a height of $h$. According to question $S_1S_2 = S_2S_3, \angle PS_2S_1 = \gamma_1, \angle PS_3S_2 = \gamma_2, \angle S_1PS_2 = \delta_1, \angle S_2PS_3 = \delta_2, \angle PS_1O = \beta_1$ and $\angle PS_2O = \beta_2$.

     In $\triangle OPS1, \sin\beta_1 = \frac{h}{PS_1} \Rightarrow PS_1 = \frac{h}{\sin\beta_1}$

     In $\triangle OPS_2, PS_2 = \frac{h}{\sin\beta_2}$

     Using sine rule in $PS_1S_2, \frac{S_1S_2}{\sin\delta_1} = \frac{PS_1}{\sin\gamma_1}$

     $\Rightarrow \frac{h}{S_1S_2} = \frac{\sin\beta_1\sin\gamma_1}{\sin\delta_1}$

     Similarly in $PS_2S_3, \frac{h}{S_2S_3} = \frac{\sin\beta_2\sin\gamma_2}{\sin\delta_2}$

     Equalting last two results we have desired equality.

181. The diagram is given below:

     

     Let $PQ$ be the vertical pillar having a height of $h$. According to question, $\tan\alpha = 2, AN = 20$ m and that $\triangle PAM$ is equilateral. Let $\angle QAP =\beta, \angle QBP = \gamma$

     In $\triangle NPQ, \tan\alpha = \frac{h}{PN} = 2 \Rightarrow PN = \frac{h}{2}$

     In $\triangle ANP, \tan60^\circ = \sqrt{3} = \frac{PN}{AN} \Rightarrow PN = 20\sqrt{3} \Rightarrow h = 40\sqrt{3}$ m.

     $\cos60^\circ = \frac{1}{2} = \frac{AN}{PA} \Rightarrow PA = 40$ m.

     $\triangle PAM$ is equilateral and $PN\perp AM \therefore AN = MN = 20$ m $\Rightarrow AM = 40$ m, $\Rightarrow AB = 80$ m.

     $\therefore PB = \sqrt{AB^2 - PA^2} = 40\sqrt{3}$ m.

     $\Rightarrow \beta = 60^\circ$ and $\gamma = 45^\circ$.

182. The diagram is given below:

     

     Let $ABC$ be the triangular park, $O$ be the mid-point of $BC$ and $OP$ be the television tower(out of the plane of paper). Given that, $\angle PAO = 45^\circ, \angle PBO = 60^\circ, \angle PCO = 60^\circ, AB = AC = 100$ m. Also, let $OP = h$ m.

     Clearly, $\angle POA = \angle POB = \angle POC = 90^\circ$.

     In $\triangle POA, \tan45^\circ = \frac{h}{OA}\Rightarrow OA = h$

     In $\triangle POB, \tan60^\circ = \frac{h}{OB}\Rightarrow OB = \frac{h}{\sqrt{3}}$

     Similarly $OC$ would be $\frac{h}{\sqrt{3}}$.

     $\because \triangle ABC$ is an isosceles triangle and $O$ is the mid-point of $BC. \therefore AO\perp BC$.

     In $\triangle AOB, AB^2 = OA^2 + OB^2 \Rightarrow h = 50\sqrt{3}$ m.

183. The diagram is given below:

     

     Let $ABCD$ be the base of the square tower whose upper corners are $A', B', C', D'$ respectively. From a point $O$ on the diagonal $AC$ the three upper corners $A', B'$ and $D'$ are visible.

     According to question $\angle AOA' = 60^\circ, \angle BOB' = \angle DOD' = 45^\circ$

     Also, $AA' = BB' = h$ and $AB = a$

     In $\triangle AA'O, \tan60^\circ = \frac{h}{AO}\Rightarrow AO = \frac{h}{\sqrt{3}}$

     In $\triangle BB'O, \tan45^\circ = 1= \frac{h}{BO} \Rightarrow BO = h$

     Using cosine rule in $\triangle AOB,$

     $\cos135^\circ = \frac{AO^2 + AB^2 - BO^2}{2AO.AB}$

     $\Rightarrow -\frac{1}{\sqrt{2}} = \frac{\frac{h^2}{3} + a^2 - h^2}{2.\frac{h}{\sqrt{3}}.a}$

     Considering $h > 0$, on simplification we arrive at $\frac{h}{a} = \frac{\sqrt{6}(1 + \sqrt{5})}{4}$.

184. The diagram is given below:

     

     In the diagram $PP'R'R$ is a plane perpendicular to the plane of the paper. Let $C$ be the center of top of the cylindrical tower. Since $A$ is the point on the horizontal plane nearest to $Q$, hence $A$ will be on the line $Q'A$ where $Q'A\perp QQ'$. According to question $QQ' = h, C'Q' = r, \angle QAQ' = 60^\circ$ and $\angle PAP' = 45^\circ$.

     In $\triangle AQQ', \tan60^\circ = \sqrt{3} = \frac{h}{AQ'} \Rightarrow AQ' = \frac{h}{\sqrt{3}}$

     In $\triangle APP', \tan45^\circ = 1 = \frac{h}{AP'} \Rightarrow AP' = h$

     $AC' = AQ' + C'Q' = \frac{h}{\sqrt{3}} + r$

     In $\triangle AC'P', AP'^2 = AC'^2 + C'P'^2 \Rightarrow h^2 + \left(\frac{h}{\sqrt{3}} + r\right)^2 + r^2$

     Taking into account that $h > 0$, on simplification we arrive at

     $\frac{h}{r} = \frac{\sqrt{3}(1 + \sqrt{5})}{2}$.

185. The diagram is given below:

     

     Let $AP$ be the pole having a height of $h$ m. Let $\angle PCA = \theta, \angle ADB = \alpha$ and $\angle BDC = \beta$. Then $\angle PBA = 2\theta$ and $\angle BPC = \theta$.

     $\Rightarrow \angle BPC = \angle BCP \Rightarrow BP = BC = 20$ m.

     From question $\tan\alpha = \frac{1}{5}, CD = 30$ m and $BC = 20$ m.

     In $\triangle BCD, \tan\beta = \frac{BC}{CD} = \frac{20}{30} = \frac{2}{3}$

     Now $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = 1$

     $\Rightarrow \alpha + \beta = 45^\circ \Rightarrow \angle ADC = \angle DAC = 45^\circ$

     $\Rightarrow AC = CD = 30$ m. $\Rightarrow AB = AC - BC = 30 - 20 = 10$ m.

     In $\triangle PAB, h^2 = PB^2 - AB^2 = 20^2 - 10^2 \Rightarrow h = 10\sqrt{3}$ m.

186. The diagram is given below:

     

     Let $OP$ be the tower having a height of $h, A$ be the initial position of the man, $B$ be the second position of the man at a distance $a$ from $A$ and $C$ be the final position of the man at a distance of $\frac{5a}{3}$ from $B$. Given that angles of elevation from $A, B$ and $C$ of the top of the tower are $30^\circ, 30^\circ$ and $60^\circ$ respectively. $OC\perp AB$ and $DN\perp OC$.

     In $\triangle POA, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{OA} \Rightarrow OA = \sqrt{3}h$

     Similarly in $\triangle POB, OB = \sqrt{3}h$ and in $\triangle POD, \tan60^\circ = \frac{h}{OD} \Rightarrow OD = \frac{h}{\sqrt{3}}$

     $\because OA = OA \Rightarrow AC = BC = \frac{a}{2}$

     $OC = \sqrt{OA^2 - AC^2} = \sqrt{3h^2 - \frac{a^2}{4}}, ON = \sqrt{OD^2 - DN^2} = \sqrt{\frac{h^2}{3} - \frac{a^2}{4}}$

     $BD = CN = OC - CN \Rightarrow \frac{5a}{3} = \sqrt{3h^2 - \frac{a^2}{4}} - \sqrt{\frac{h^2}{3} - \frac{a^2}{4}}$

     On simplification, we get $h = \sqrt{\frac{85}{48}}a$ or $h = \sqrt{\frac{5}{6}}a$.

187. The diagram is given below:

     

     Let $OP$ be the tower having a height of $h$. Given $ABC$ is an equilateral triangle. Let the angle subtended by $OP$ at $A, B, C$ be $\alpha, \beta, \gamma$ respectively. According to question $\tan\alpha = \sqrt{3} + 1, \tan\beta = \sqrt{2}$ and $\tan\gamma = \sqrt{2}$. $OP$ is perpedicular to the plane of $\triangle ABC$.

     In $\triangle AOP, \tan\alpha = \frac{h}{OA}\Rightarrow OA = \frac{h}{\sqrt{3} + 1}$.

     Similarly, $OB = \frac{h}{\sqrt{2}}$ and $OC = \frac{h}{\sqrt{2}}$.

     In $\triangle AOB$ and $AOC, AB = AC, OB = OC$ and $OA$ is common. So $\triangle AOB$ and $\triangle AOC$ are equal. $\therefore \angle OAB = \angle OAC$.

     But $\angle BAC = 60^\circ \therefore \angle OAB = \angle OAC = 30^\circ$

     Let $\angle OBA = \theta$

     Using sine rule in $\triangle OAB, \frac{OB}{\sin30^\circ} = \frac{OA}{\sin\theta}$

     $\Rightarrow \sin\theta = \frac{\sqrt{3} - 1}{2\sqrt{2}} = \sin15^\circ$

     $\Rightarrow \theta = 15^\circ. \Rightarrow \angle OBD = \angle ABC - \theta = 45^\circ$

     In $\triangle BOC, OB = OC, OD\perp BC \therefore BD = DC = 40'$

     In $\triangle BOD, \cos45^\circ = \frac{BD}{OB} = \frac{40}{h/\sqrt{2}}\Rightarrow h = 80'$

188. The diagram is given below:

     

     Let $OP$ be the tower having a height of $h$ and $PQ$ be the flag-staff having a height of $x$. Since $PQ$ subtends equal angle $\alpha$ at $A$ and $B$ so a circle will pass through $A, B, P$ and $Q$. Since $C$ is the mid-point of $AB \therefore AC = BC = a$.

     Let $OA = d$ and $\angle PAO = \theta$. In $\triangle AOP, \tan\theta = \frac{h}{d}$

     In $\triangle AOQ, \tan(\theta + \alpha) = \frac{h + x}{d} \Rightarrow \frac{\tan\theta + \tan\alpha}{1 - \tan\theta\tan\alpha} = \frac{\frac{h}{d} + \tan\alpha}{1 - \frac{h}{d}\tan\alpha}$

     $\Rightarrow \frac{h + d\tan\alpha}{d - h\tan\alpha} = \frac{h + x}{d}$

     $\Rightarrow d^2 + h(x + h) = xd\cot\alpha$

     Similarly, $(d + a)^2 + h(x + h) = x(d + a)\cot\beta$

     As the points $A, B, P$ and $Q$ are concyclic $\therefore OA.OB = OP.OQ$

     $d(d + 2a) = h(h + x)$

     $\Rightarrow d^2 + d(d + 2a) = xd\cot\alpha \Rightarrow d + a = \frac{x}{2}\cot\alpha$

     Similarly, $(d + a)^2 + (d + a)^2 - a^2 = x(d + a)\cot\beta$

     Solving the above two equations

     $\frac{x^2}{4}\cot^2\alpha + \frac{x^2}{4}\cot^2\alpha - a^2 = x.\frac{x}{2}\cot\alpha\cot\beta$

     $\Rightarrow \frac{x^2}{2}(\cot^2\alpha - \cot\alpha\cot\beta) = a^2$

     $x = a\sin\alpha\sqrt{\frac{2\sin\beta}{\cos\alpha\sin(\beta - \alpha)}}$

189. The diagram is given below:

     

     Let $A_1, A_2, \ldots, A_{10}, \ldots, A_{17}$ be the feet of the first, second, …, tenth, and seventeenth pillars respectively and $h$ be the height of each of these pillars. Given that these pillars are equidistant, therefore $A_1A_2 = A_2A_3 = \cdots = A_{16}A_{17} = x$ (let).

     Clearly, $A_1A_{10} = 9x$ and $A_1A_{17} = 16x$. We have let $O$ as the position of the observer and $\angle A_2A_1O = \theta$.

     In $\triangle A_{10}OP, \tan\alpha = \frac{h}{OA_{10}}\Rightarrow OA_{10} = h\cot\alpha$

     Similarly, $OA_{17} = h\cot\beta$

     From question $OA_1 = \frac{h\cot\alpha}{2}$ and $OA_1 = \frac{h\cot\beta}{3}$

     $\Rightarrow 2OA1 = OA_{10}$ and $3OA_1 = OA_{17}$. Let $OA_1 = y$ then

     $OA_{10} = 2y$ and $OA_{17} = 3y$

     Using cosine rule in $\triangle OA_1A_{10}, \cos\theta = \frac{81x^2 + y^2 - 4y^2}{2.9x.y}$

     $\Rightarrow y^2 = 27x^2 - 6xy\cos\theta$

     Similarly in $\triangle OA_1A_{17}, y^2 = 32x^2 - 4xy\cos\theta$

     $\Rightarrow y^2 = 42x^2 \Rightarrow \frac{y}{x} = \sqrt{42}$

     $\Rightarrow \sec\theta = -\frac{2\sqrt{42}}{5}$

     Acute angle will be given by $\sec\theta = \left|-\frac{2\sqrt{42}}{5}\right| = 2.6$ (approximately).

190. The diagram is given below:

     

     Let $DP$ be the tower having a height of $h$ with foot at $D$ and $A, B, C$ be the three points on the ciircular lake. According to question $\angle PAD = \alpha, \angle PBD = \beta$ and $\angle PCD = \gamma$. Also, $\angle BAC = \theta$ and $\angle ACB = \theta$. We know that angles on the same segement of a circle are equal. $\therefore \angle ADB = \angle ACB = \theta$ and $\angle BDC = \angle BAC = \theta$.

     In $\triangle PDA, \tan\alpha = \frac{h}{AD} \Rightarrow AD = h\cot\alpha$

     Similarly, $BD = h\cot\beta$ and $CD = h\cot\gamma$

     In $\triangle ABC, \angle BAC = \angle ACB \Rightarrow AB = BC \Rightarrow AB^2 = bC^2$

     Using cosine rule in $\triangle ABD, \cos\theta = \frac{AD^2 + BD^2 - AB^2}{2.AD.BD}$

     $\Rightarrow AB^2 = AD^2 + BD^2 - 2.AD.BD.\cos\theta$

     Similarly in $\triangle BDC, BC^2 = BD^2 + CD^2 - 2.BD.CD.\cos\theta$

     $\Rightarrow AD^2 + BD^2 - 2.AD.BD.\cos\theta = BD^2 + CD^2 - 2.BD.CD.\cos\theta$

     $\Rightarrow 2.BD.\cos\theta[CD - AD] = CD^2 - AD^2$

     $\Rightarrow 2.BD.\cos\theta = CD + AD \Rightarrow 2\cos\theta\cot\beta = \cot\alpha + \cot\gamma$.

191. The diagram is given below:

     

     Let $DP$ be the pole of height $h$ and $R$ be the radius of the circular pond. According to question, $\angle PAD = \angle PBD = 30^\circ$ and $\angle PCD = 45^\circ$.

     Clearly, $\angle PDA = \angle PDB = \angle PDC = 90^\circ$

     Also arc $AB = 40$ m and arc $BC = 20$ m.

     Now $\frac{2\pi R}{40} = \frac{2\pi}{\angle AOB}\Rightarrow \angle AOB = \frac{40}{R}$

     Similarly, $\angle BOC = \frac{20}{R} \therefore \angle AOB = 2\angle BOC$

     $\Rightarrow \angle ADB = 2.\angle BDC$ [$\because$ angle subtended by a segment at the center is double the angle subtended at circumference.]

     Let $\angle BDC = \theta$, then $\angle ADB = 2\theta$.

     In $\triangle PDA, \tan30^\circ = \frac{h}{AD} \therefore AD = \sqrt{3}h$.

     Similarly, in $\triangle PDB, BD = \sqrt{3}h$ and in $\triangle PDC, CD = h$.

     Now $\because AD = BD \therefore \angle DAB = \angle DBA = 90^\circ - \theta$

     Also, $\angle BAC = \angle BDC = \theta$ and $\angle ACB = \angle ADB = 2\theta$.

     Now $\angle ABC = 180^\circ - 3\theta \therefore \angle DBC = \angle ABC - \angle ABD$

     $= (180^\circ - 3\theta) - (90^\circ - \theta) = 90^\circ - 2\theta$

     $\angle BCD = 90^\circ + \theta$

     Using sine rule in $\triangle BCD, \frac{BD}{sin\angle BCD} = \frac{CD}{sin\angle DBC}$

     $\Rightarrow \frac{\sqrt{3}h}{\sin(90^\circ + \theta)} = \frac{h}{\sin(90^\circ - 2\theta)}$

     $\Rightarrow \frac{\sqrt{3}}{\cos\theta} = \frac{1}{\cos2\theta}$

     $\Rightarrow \cos\theta = \frac{\sqrt{3}}{2}, -\frac{1}{\sqrt{3}}$ (rejected because $\theta \ngtr 90^\circ$)

     $\theta = 30^\circ \Rightarrow \angle ADC = 3\theta = 90^\circ$

     $\therefore AC$ will be the diameter. arc $ABC =$ semiperimeter $= 60$ m.

     $\pi R = 60 \Rightarrow R = 19.09$ m.

     In $\triangle ADC, AC^2 = AD^2 + CD^2 \Rightarrow 4R^2 = 3h^2 + h^2 \Rightarrow h = R = 19.09$ m.

192. The diagram is given below:

     

     Let the man start at $O$ on the straight sea shore $OAB, P$ and $Q$ be the buoys. According to question, $OA = a, OB = b, \angle POA = \alpha - \angle PAQ = \angle PBQ$.

     $\because\angle PAQ = \angle PBQ = \alpha \therefore$ a circle will pass through the points $A, B, P$ and $Q$.

     Let $\angle OAQ = \theta \angle QAB = \pi - \theta$

     Also, $\angle OQA = \pi - (\angle QOA + \angle OAQ) = \pi - (\alpha + \theta)$

     $\therefore \angle APQ = \pi - (\angle PAQ + \angle PQA) = \pi - [\alpha + \pi - (\alpha + \theta)] = \theta$

     Since $ABPQ$ is concyclic $\therefore \angle ABQ = \pi - \angle APQ = \pi - \theta = \angle QAB \Rightarrow QA = QA \therefore \triangle QAB$ is an isosceles triangle.

     Draw $OD\perp AB \therefore D$ is the mid-point of $AB$.

     $\Rightarrow AD = BD = \frac{b}{2}\Rightarrow OD = OA + AD = a + \frac{b}{2}$

     In $\triangle ODQ, \cos\alpha = \frac{OD}{OQ} \Rightarrow OQ = \left(a + \frac{b}{2}\right)\sec\alpha$

     From the properties of a circle, $OA.OB = OP.OQ$

     $\Rightarrow a.(a + b) = OP.\left(a + \frac{b}{2}\right)\sec\alpha$

     $\Rightarrow OP = \frac{2a(a + b)\cos\alpha}{2a + b}$

     $\Rightarrow PQ = OQ - OP = \left(a + \frac{b}{2}\right)\sec\alpha - \frac{2a(a + b)}{2a + b}\cos\alpha$.

193. The diagram is given below:

     

     Let $A_1OA_n$ be the railway curve in the shape of a quadrant, the telegraph posts be represented by $A_1, A_2, \ldots, A_n$ and the man be stationed at $C$. From question $CPQ$ is a straight line. Also, $A_1C = a$. Let $OA_1$ be the radius of the quadrant and $O$ its center. Clearly, $A_1OA_n = \frac{\pi}{2}$.

     As there are $n$ telegraph posts from $A_1$ to $A_n$ at equal distances, arc $A_1A_N$ is divided in $n - 1$ equal parts.

     $\therefore \angle A_1OA_2 = \angle A_2OA_3 = \cdots = A_{n - 1}OA_n = \frac{\pi}{2(n - 1)} = \theta$

     According to question, $\phi = \frac{\pi}{4(n - 1)} \Rightarrow \theta = 2\phi$.

     Let $P$ and $Q$ be the $p^{th}$ and $q^{th}$ posts as seen from $A_1$.

     $\therefore \angle A_1OP = p\theta = 2p\phi$ and $\angle A_1OQ = q\theta = 2q\phi$

     $\angle POQ = (q - p)\theta = 2(q - p)\phi$. Draw $OD\perp PQ$

     $\because OP = OQ =$ radius of the circular quadrant. $\therefore \triangle POQ$ is an isosceles triangle.

     Clearly, $OD$ bisects the $\angle POQ \therefore \angle POD = \angle QOD = (q - p)\phi$

     $\angle COD = \angle A_1OD = \angle A_1OP + \angle POD = 2p\phi + (q - p)\phi = (p + q)\phi$

     In $\triangle ODC, \cos\angle COD = \frac{OD}{OC} \Rightarrow \cos(p + q)\phi = \frac{OD}{r + a}$

     $\Rightarrow OD = (r + a)\cos(p + q)\phi$

     In $\triangle ODP, \cos\angle POD = \frac{OD}{OP} \Rightarrow \cos(p - q)\phi = \frac{OD}{r}\Rightarrow OD = r\cos(q - p)\phi$

     $\Rightarrow (r + a)\cos(q + p)\phi = r\cos(q - p)\phi$

     $\Rightarrow -a\cos(q + p)\phi = r[\cos(q + p)\phi - \cos(q - p)\phi]$

     $\Rightarrow r = \frac{a}{2}\cos(q + p)\phi.\cosec p\phi.\cosec q\phi$.

194. The diagram is given below:

     

     Let $r$ be the radius of the wheel and $x$ be the length of the rod. Clearly, $AC = 2r + x$. According to question $\angle APC = \alpha$.

     In $\triangle PAC, \tan\alpha = \frac{AC}{AP} = \frac{2r + x}{d} \Rightarrow x = d\tan\alpha - 2r$.

     After rotation of the wheel, let $C'$ be the new position of $C$ as shown in the figure. In this case angle of elevation of $C'$ is $\beta$. Since $C'$ is the position of $C$ when it is about to disappear, so $PC'$ will be tangent to the wheel. Let it touch the wheel at $Q$.

     In $\triangle OPQ$ and $APO, OQ = OA = r, OP$ is common.

     $\angle OQP = \angle OAP = 90^\circ \therefore$ triangle are equal.

     $\Rightarrow \angle OPQ = \angle OPA = \frac{\beta}{2}$

     In $\triangle OAP, \tan\frac{\beta}{2} = \frac{OA}{AP} = \frac{r}{d} \Rightarrow r = d\tan\frac{\beta}{2}$

     $\Rightarrow x = d\left(\tan\alpha - 2\tan\frac{\beta}{2}\right)$

     $PA$ and $PQ$ are tangents to the same circle $\therefore PQ = PA = d$

     $\therefore \angle OQC' = 90^\circ$

     In $\triangle OQC', QC' = \sqrt{OC'^2 - OQ^2} = \sqrt{(x + r)^2 - r^2} = \sqrt{x(x + 2r)}$

     $= d\sqrt{\tan^2\alpha - 2\tan\alpha\tan\frac{\beta}{2}}$

     $\therefore PC' = PQ + QC' = d + d\sqrt{\tan^2\alpha - 2\tan\alpha\tan\frac{\beta}{2}}$

195. The diagram is given below:

     

     Let $PQ$ be the tower having a height of $h, ADB$ be the arc having the given length of $2L$ and $AC$ be the part of arc with length $\frac{L}{2}$. Clearly, line $PC$ will be tangent to the arch as the man at $C$ just sees the topmost point $P$ of the tower. $D$ is the topmost point of the semi-circullar arch.

     Let $r$ be the radius of the arch. According to question $\angle PDT = \theta$ where $DT\perp PQ$.

     Let $\angle COA = \phi$. Here $O$ is the center of the arch. Clearly, $2L$ length represents semi-cicular arch which means $AC$ which is of length $\frac{L}{2}$ will make an angle of $45^\circ$ at center i.e. $\phi = 45^\circ$.

     In $\triangle ONC, CN = OC\sin\phi$ and $ON = OC\cos\phi$

     $\Rightarrow CN = \frac{r}{\sqrt{2}}$ and $ON = \frac{r}{\sqrt{2}}$

     Let $CR\perp PQ$ then $CD\parallel NO \therefore \angle OCM = \angle CON = 45^\circ$

     Also, $\angle OCP = 90^\circ$ beccause $OC$ is normal at $C$.

     $\therefore \angle PCR = \angle PCO - \angle OCR = 90^\circ - 45^\circ = 45^\circ$

     In $\triangle PRC, \tan45^\circ = \frac{PR}{CR} = \frac{PQ - QR}{CR} = \frac{h - \frac{r}{\sqrt{2}}}{CR}$

     $\Rightarrow CR = h - \frac{r}{\sqrt{2}}$

     In $\triangle DPT, \tan\theta = \frac{PT}{DT} = \frac{PQ - QT}{MR} = \frac{PQ - OD}{CR - CM}$

     $\Rightarrow \tan\theta = \frac{h - r}{h- \frac{r}{\sqrt{2}} - \frac{h}{\sqrt{2}}}$

     $\Rightarrow h = \frac{r(\sqrt{2}\tan\theta - 1)}{\tan\theta - 1} = \frac{2L}{\pi}.\frac{\sqrt{2}\tan\theta - 1}{\tan\theta - 1}$.

196. The diagram is given below:

     

     According to question, $\angle DAB = \alpha, \angle CAB = \beta$

     $\therefore \angle CAD = \beta - \alpha.\because AC$ is the diameter. $\therefore \angle ABC = 90^\circ$. Let $O$ be the ceter of the circle and $r$ be its radius then $AC = 2r$.

     $\because E$ is the mid-point of $CD. \therefore CE = ED = x$ (let)

     $\because \angle ADC$ is the exterior angle of $\triangle ABC. \therefore \angle ADC = 90^\circ + \alpha$

     Using sine rule in $\triangle ADC, \frac{2r}{\sin(90^\circ + \alpha)} = \frac{2x}{\sin(\beta - \alpha)} \Rightarrow x = \frac{r\sin(\beta - \alpha)}{\cos\alpha}$

     In $\triangle ABC, \cos\alpha = \frac{AB}{AD} = \frac{2r\cos\beta}{AD}\Rightarrow AD = \frac{2r\cos\beta}{\cos\alpha}$

     $\because AE$ is the median of the $\triangle CAD. \therefore AC^2 + AD^2 = 2(AE^2 + CE^2)$

     $\Rightarrow 4r^2 + \frac{4r^2\cos^2\beta}{\cos^2\alpha} = 2d^2 + 2x^2$

     $\Rightarrow \frac{4r^2(\cos^2\alpha + \cos^2\beta)}{\cos^2\alpha} = 2d^2 + \frac{r^2\sin^2(\beta - \alpha)}{\cos^2\alpha}$

     $\Rightarrow r^2 = \frac{d^2\cos^2\alpha}{2\cos^2\alpha + 2\cos^2\beta - \sin^2(\beta - \alpha)}$

     $\Rightarrow = \frac{d^2\cos^2\alpha}{\cos^2\alpha + \cos^2\beta + \cos(\alpha + \beta)\cos(\beta - \alpha) + \cos^2(\beta - \alpha)}$

     $= \frac{d^2\cos^2\alpha}{\cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta\cos(\beta - \alpha)}$

     Thus are of the triangle can be found which is equal to desired result.

197. The diagram is given below:

     

     Let $AB$ be the surface of the lake and $C$ be the point of observation such that $AC = h$ m. Let $E$ be the position of the cloud and $E'$ be its reflection then $BE = BE'$.

     In $\triangle CDE, \tan\alpha = \frac{DE}{CD} = \frac{H}{CD} \Rightarrow CD = \frac{H}{\tan\alpha}$

     In $\triangle CDE', \tan\beta = \frac{DE'}{CD} = \frac{2h + H}{CD} \Rightarrow CD = \frac{2h + H}{\tan\beta}$

     $\Rightarrow \frac{H}{\tan\alpha} = \frac{2h + H}{\tan\beta} \Rightarrow H(\tan\beta - \tan\alpha) = 2h\tan\alpha \Rightarrow H = \frac{2h\tan\alpha}{\tan\beta - \tan\alpha}$

     In $\triangle CDE, \sin\alpha = \frac{DE}{CE} \Rightarrow CE = \frac{H}{\sin\alpha}$

     $\Rightarrow CD = \frac{2h\tan\alpha}{(\tan\beta - \tan\alpha)\sin\alpha} = \frac{2h\sec\alpha}{\tan\beta - \tan\alpha}$

     $= \frac{2h\sec\alpha.\cos\alpha.\cos\beta}{\sin\beta\cos\alpha - \sin\alpha\cos\beta}$

     $= \frac{2h\cos\beta}{\sin(\beta - \alpha)}$.

198. Front view and side view are given below:

     

     In $\triangle ADE, \tan30^\circ = \frac{AD}{DE} \Rightarrow DE = \sqrt{3}h$

     In $\triangle BDE, \tan\alpha = \frac{BD}{DE} = \frac{a}{\sqrt{3}h}$

     Tangent of apex of shadow $= \tan2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}$

     $= \frac{\frac{2a}{\sqrt{3}h}}{1 - \frac{a^2}{3h^2}} = \frac{2ah\sqrt{3}}{3h^2 - a^2}$.

199. Let $ABCD$ be the target and $ABC'D'$ be its shadow then $\angle DAD' = \beta$ and $\angle BAD'= 90^\circ - \beta$. Area of the target $= AB.AD$ and area of the shadow $= AB.AD'.\cos\beta$

     $\frac{\text{Area of target}}{\text{Area of shadow}} = \frac{AB.AD}{AB.AD'.\cos\beta} = \tan\alpha.\sec\beta$

200. This question is similar to $169$, and has been left as an exercise.