# 31. Height and Distance Solutions Part 4#

1. 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

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

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

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

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

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

3. 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.

4. 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.

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

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

6. 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.

7. 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 $\triangle BCE, \tan\beta = \frac{h}{d + h\cot\theta}$

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

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

8. 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.

9. 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.

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

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

12. 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 $\triangle ACD$,

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

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

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

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

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

16. 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.

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

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

18. 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 $\triangle O'DE$,

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

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

19. 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 $\triangle OML$,

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

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

20. 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 $\triangle ABQ, \frac{3d}{4\sin\theta} = \frac{d}{\sin(180^\circ - \theta)}$

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

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

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

21. 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, $\tan\beta = \frac{1}{8}$

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

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

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

23. 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.

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

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

25. 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.

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

26. 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,

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

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

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

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

28. 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.

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

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

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

29. 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.

30. 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.

31. 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 $\gamma = 45^\circ$.

32. 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.

33. 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 $\frac{h}{a} = \frac{\sqrt{6}(1 + \sqrt{5})}{4}$.

34. 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

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

35. 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.

36. 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 $h = \sqrt{\frac{5}{6}}a$.

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

38. 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

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

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

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

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