32. Height and Distance Solutions Part 5#

1. The diagram is given below:

Let $BC$ be the tower having a height of $h$. According to question $AB = h$ and $BD = h/2$.

In $\triangle BCD, \tan\alpha = \frac{h}{\frac{h}{2}} = 2$

In $\triangle ABC, \tan\beta = \frac{h}{h} \Rightarrow \alpha = 45^\circ$

$L\tan\alpha = 10 + \log 2 = 10.30103$ (given)

$L\tan\alpha - L\tan63^\circ26'= 10.30103 - 10.30094$

If the difference is $x^2$, then $x^2 = \frac{60\times306}{3152} = 5.81^2$

$\therefore \alpha = 63^\circ26'6''$

Change in sun’s altitude $= 63^\circ26'6'' - 45^\circ = 18^\circ26'6''$

2. This problem is similar to $117$, and has been left as an exercise.

3. Since the angle of elevation is same from every point it is clear that the balloon is above the circumcenter. From sine rule, $\sin\beta = \frac{b}{2R} \Rightarrow R = \frac{b}{2\sin\beta}$.

If $h$ is the height of the balloon then $\tan\alpha = \frac{h}{R} \Rightarrow h = R\tan\alpha$

$= \frac{b}{2}\tan\alpha.\cosec\beta$

4. The diagram is given below:

In the diagram $A$ representes the lighthouse. $AB$ is north-east direction and $AC$ is north-west direction between which the light is thrown by the lighthouse. Let $BC$ is the westward path of the steamer. Given that the distance of steamer is $5$ k m from lighthouse i.e. $AB = 5$ km and clearly $\triangle ABC$ is a isosceles, right angled triangle so $AC = 5$ km and $BC = \sqrt{AB^2 + AC^2} = \sqrt{5^2 + 5^2} = 5\sqrt{2}$ km.

Speed of steamer $= \frac{5\sqrt{2}}{30\sqrt{2}} = \frac{1}{6}$ km/min.

5. The diagram is given below:

Let $A$ be the initial position of the man and $D$ be the initial position of the balloon with angle of elevation equal to $60^\circ$. The man moves northward for $400$ yards and balloon moves north-west and they meet at $C$ where the balloon is vertically above the man. Since $BC$ is north-west the $\triangle ABC$ will be isosceles, right angled triangle with $AB = AC = 400$ yards.

$\therefore$ in $\triangle ABC, \tan60^\circ = \frac{h}{400} \Rightarrow h = 400\sqrt{3}$ yards.

6. The diagram is given below:

Let $ABCD$ be the vertical cross-section of the tower through the middle, let the side of the square is $AB$ having length $a$ and height of the tower be $OP$ equal to $h$. Let the height of flag-staff $PQ$ be $b$. $M$ and $N$ are points of observation such that $AN = MN = 100$ m. Let $\alpha$ and $\beta$ be angles of elevation from $M$ of $D$ and $Q$ such that $\tan\alpha = \frac{5}{9}$ and $\tan\beta = \frac{1}{2}$. At $N$ the man just sees the flag.

$\tan\beta = \frac{1}{2} = \frac{AD}{AM} = \frac{AD}{100} \Rightarrow AD = 100 = OP = h$.

$\therefore AD = AN = 100 \Rightarrow \angle AND = \angle NDA = 45^\circ \Rightarrow PD = PQ \Rightarrow \frac{a}{2} = b \Rightarrow a = 2b$

$\tan\alpha = \frac{5}{9} = \frac{OQ}{OM} = \frac{b + h}{200 + \frac{a}{2}} = \frac{b + h}{200 + b}$

$\Rightarrow b = 25 \Rightarrow a = 50$.

7. The diagram is given below:

Let $OC$ be the vertical pole having a height of $h$. $A$ and $B$ are given points in the question from where anglea of elevations of $C$ are $\alpha$ and $\beta$ respectively. Angle subtended by $AB$ at $O$ is $\gamma$ as shown in the diagram. Let $OB = x$ and $OA = y$. Given that $AB = d$

In $\triangle OAC, \tan\alpha = \frac{h}{y}\Rightarrow y = h\cot\alpha$

Similarly $x = h\cot\beta$

In $\triangle OAB, d^2 = x^2 + y^2 - 2xy\cos\gamma$

$d^2 = h^2\cot^2\alpha + h^2\cot^2\beta - 2h^2\cot\alpha\cot\beta\cos\gamma$

$\Rightarrow h = \frac{d}{\sqrt{\cot^2\alpha + \cot^2\beta - 2\cot\alpha\cot\beta\cos\gamma}}$

8. The diagram is given below:

Let $OP$ be the tree having a height of $h$ and $OAB$ is the hill inclines at angle $\alpha$ with the horizontal. Let $A$ be the point from where angle of elevation of the top of the tree be $\beta$ and $B$ be the point from where the angle of depression of the top of the tee be $\gamma$. Given $AB = m$.

$\angle POA = 90^\circ - \alpha, \angle OAP = \alpha + \beta, \angle = \alpha - \gamma$ and $\angle ABP = (\alpha + \beta) - (\alpha - \gamma) = \beta + \gamma$

In $\triangle OAP, \frac{AP}{\sin(90^\circ - \alpha)} = \frac{OP}{\sin(\alpha + \beta)}$

$\Rightarrow \frac{AP}{\cos\alpha} = \frac{h}{\sin(\alpha + \beta)}$

In $\triangle PAB, \frac{AB}{\sin(\beta + \gamma) = \frac{AP}{\sin(\alpha - \gamma)}}$

$\Rightarrow h = \frac{m\sin(\alpha + \beta)\sin(\alpha - \gamma)}{\cos\alpha\sin(\beta + \gamma)}$

9. The diagram is given below:

Let $ABCDA'B'C'D'$ be the vertical tower having a height of $h$ i.e. $AA' = BB' = CC' = DD' = h$ with side equal to $b$ and $O$ be the point of observation on the diagonal $AC$ extended at a distance $2a$ from $A$. Clearly, $\angle OAB = 135^\circ$. Given that $\angle AOA' = 45^\circ$ and $\angle BOB' = 30^\circ$.

In $\triangle AOA', \tan45^\circ = 1 = \frac{AA'}{OA} = \frac{h}{2a}\Rightarrow h = 2a$

In $\triangle BOB', \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BB'}{OB} \Rightarrow OB = 2\sqrt{3}a$

In $\triangle OAB, \cos135^\circ = -\frac{1}{\sqrt{2}} = \frac{4a^2 + b^2 - 12a^2}{2.2a.b}$

This is a quadratic equation in b wiht two roots $a(-\sqrt{2}\pm\sqrt{10})$. Clearly, $b$ cannot be negaitive so $b = a(\sqrt{10} - \sqrt{2})$.

10. The diagram is given below:

Let $AS$ be the steeple having a height of $h, B$ is the point due south having an angle of elevation of $45^\circ$ to the top of the tower and $C$ is the point due south of $B$, at a distance of $a$ from $B$, having an angle of elevation of $15^\circ$ to the top of the tower. $AS$ is perpendicular to the plane of paper.

In $\triangle ABS, \tan45^\circ = 1 = \frac{AS}{AB}\Rightarrow AB = h$

In $\triangle ACS, \tan15^\circ = 2 - \sqrt{3} = \frac{AS}{AC} \Rightarrow AC = h(2 + \sqrt{3})$

In $\triangle ABC, AC^2 = AB^2 + BC^2 \Rightarrow h^2(2 + \sqrt{3})^2 = h^2 + a^2$

$\Rightarrow a = \frac{h}{\sqrt{6 + 2\sqrt{3}}}$

11. The diagram is given below:

Let $CD$ be the given tower with a given height of $c$, $FG$ be the mountain behind the spire and tower at a distance $x$ having a height of $h$ and $A$ and $B$ are the points of observation. Let the angle of elevation from $A$ is $\alpha$ such that the mountain is just visible behind the tower. Let $DE$ be the spire which subtends equal angle of $\beta$ at $A$ and $B$. Since it subtends equal angles at $A$ and $B$ the points $A, B, D$ and $E$ will be concyclic. $a$ and $b$ are shown as given in the question.

$\angle AEC = 90^\circ - (\alpha + \beta)$

Segment $AD$ will also subtend equal angles at $B$ and $E$ $\therefore \angle AED = \angle ABD = 90^\circ - (\alpha + \beta) \Rightarrow \angle CBE = 90^\circ - \alpha$

In $\triangle ACD$ and $AFG, \tan\alpha = \frac{c}{a} = \frac{h}{x + a} \Rightarrow x = \frac{ah - ac}{c}$

In $\triangle BFG, \tan(90^\circ - \alpha) = \frac{h}{x + a + b}$

$\Rightarrow \frac{a}{c} = \frac{h}{\frac{ah - ac}{c} + a + b}$

$\Rightarrow h = \frac{abc}{c^2 - a^2}$

12. The diagram is given below:

Let $AB$ be the pole having height $h$ and $CD$ be the tower having a height of $h + x$ as shown in the diagram. The angles $\alpha$ and $\beta$ are shown as given in the question. Let $d$ be the distance between the pole and tower. Clearly, $\angle ADC = 90^\circ - \alpha \Rightarrow \angle BDC = 90^\circ - (\alpha - \beta)$. Let $h + x = H$

In $\triangle ACD, \tan\alpha = \frac{h + x}{d} \Rightarrow d = (h + x)\cot\alpha = H\cot\alpha$

In $\triangle BDE, \tan(\alpha - \beta) = \frac{H - h}{d} \Rightarrow d = (H - h)\cot(\alpha - \beta)$

$\Rightarrow H\cot\alpha = (H - h)\cot(\alpha - \beta)$

$\Rightarrow H = \frac{h\cot(\alpha - \beta)}{\cot(\alpha- \beta) - \cot\alpha}$

13. The diagram is given below:

Let $A, B, C$ and $D$ be the points on one bank such that $AB = 6d, AC = 2d, AD = BD = 3d$ and $PQ$ be the tower on the other bank perpendicular to the plane of the paper having a height of $h$. Given that $\angle PBQ = \angle PAQ = \alpha$ and $\angle PCQ = \beta$.

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

Similarly, $PA = h\cot\alpha$ and $PC = h\cot\beta$. Since $PA = PA$ the $\triangle PAB$ is an isosceles triangle. As $D$ is the mid-point of $AB$ so $\triangle PBD, \triangle PCD$ and $\triangle PAD$ will be right angled triangles.

In $\triangle PAD, PA^2 = PD^2 + AD^2$ and in $\triangle PCD, PC^2 = PD^2 + CD^2$

Subtracting, we get $PA^2 - PC^2 = AD^2 - CD^2$

$\Rightarrow h^2(\cot^2\alpha - \cot^2\beta) = 9d^2 - d^2 = 8d^2$

$\Rightarrow h = \frac{2\sqrt{2}d}{\sqrt{\cot^2\alpha - \cot^2\beta}}$

$PD$ represents the width of the canal. $\Rightarrow PD^2 = PA^2 - AD^2 = h^2\cot^2\alpha - 9d^2$

$\Rightarrow PD = d\sqrt{\frac{9\cot^2\beta - \cot^2\alpha}{\cot^2\alpha - \cot^2\beta}}$

14. The diagram is given below:

Let $PQ$ be the tower having a height of $h$ and points $A, B$ are the two stations at a distance of $2$ km having angles of elevation of $60^\circ$ and $30^\circ$ respectively. $C$ is the mid-point between $A$ and $B$ from where the angle of elevation is $45^\circ$.

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

Similarly, $PA = \sqrt{3}h$ and $PC = h$.

Now since $C$ is the mid-point of $AB$ therefore $PC$ is the median of the triangle $PAB$.

$\Rightarrow PA^2 + PB^2 = 2(PC^2 + AC^2)$

$\Rightarrow \frac{h^2}{3} + 3h^2 = 2(h^2 + 1)$

$\Rightarrow h = \frac{\sqrt{3}}{\sqrt{2}}$ km $= 500\sqrt{6}$ m.

15. The diagram is given below:

Let $PQ$ be the flag-staff standing inside equilateral $\triangle ABC$ and since all sides subtend an angle of $60^\circ$ it is guaranteed that $P$ will be centroid of the $\triangle ABC$. Given that the height of the flag-staff is $10$ m. Also, according to question $\angle AQB = \angle BQC = \angle CQA = 60^\circ \therefore AQ = BQ$. Let each side of the triangle has length of $2a$ m.

Thus, $\triangle AQB$ is an equilateral triangle. $\therefore AQ = BQ = AB = 2a = CQ$

We know from geometry that $AP = \frac{2}{3}AD$. We also know that median of an equilateral triangle is perpendicular bisector. $\therefore \triangle ABD$ is a right-angle triangle where $D$ is the point where $AP$ would meet $BC$.

$\Rightarrow \sin60^\circ = \frac{AD}{AB} \Rightarrow AD = 2a\sin60^\circ$

$\Rightarrow AP = \frac{2a}{\sqrt{3}}$

$\triangle APQ$ is also a right angle triangle.

$\Rightarrow AQ^2 = AP^2 + PQ^2 \Rightarrow 4a^2 - \frac{4a^2}{3} = 10$

$\Rightarrow a = 5\sqrt{\frac{3}{2}}$

$\Rightarrow 2a = 5\sqrt{6}$ m.

16. The diagram is given below:

Let $CD$ be the cliff having a height of $H, DE$ be the tower on the cliff having a height of $h$ and $A, B$ are two points on horizontal level where the tower subtends the equal angle $\beta$ at a distance of $a, b$ from the cliff’s foot. Let $\alpha$ be the angle of elevation from $A$ of the cliff’s top.

Since $DE$ subtends equal angles at $A, B$ therefore a circle will pass through these four points and thus chord $AD$ will also subtend equal angles $\angle AEC$ and $\angle ABD$ equal to $90^\circ - (\alpha + \beta)$.

In $\triangle ACD, \tan\alpha = \frac{H}{a}$ and $\tan(\alpha + \beta) = \frac{H + h}{a} = \frac{b}{H}$

In $\triangle BCE, \tan(90^\circ - \alpha) = \cot\alpha = \frac{H + h}{b}$

We have $\frac{H + h}{a} = \frac{b}{H} \Rightarrow ab - H^2 = Hh$

We have $\tan(\alpha + \beta) = \frac{b}{H}$

$\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{b}{H}$

$\Rightarrow \frac{\frac{H}{a} + \tan\beta}{1 - \frac{H}{a}\tan\beta} = \frac{b}{H}$

$\Rightarrow \left(H + \frac{bH}{a}\right)\tan\beta = \frac{ab - H^2}{a}$

$\Rightarrow h = (a + b)\tan\beta$

17. The diagram is given below:

Let $AB$ be the tower and $BC$ be the flag-staff having heights of $x$ and $y$ respectively. According to question $BC$ makes an angle of $\alpha$ at $E$ which is $c$ distance from the tower. Let the angle of elevation from $E$ to the top of tower $B$ is $\beta$.

In $\triangle ABE, \tan\beta = \frac{x}{c}$

In $\triangle ACE, \tan(\alpha + \beta) = \frac{x + y}{c}$

$\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{x + y}{c}$

$\Rightarrow \frac{x + c\tan\alpha}{c - x\tan\alpha} = \frac{x + y}{c}$

$\Rightarrow \tan\alpha = \frac{cy}{x^2 + c^2 + xy}$

Given that $\alpha$ is the greatest angle made which means $\tan\alpha$ will be greatest. So equating the derivative w.r.t to $c$ to zero, we get

$\frac{d}{dc}\left[\frac{cy}{x^2 + c^2 + xy}\right] = \frac{c[x(x + c) - c^2]}{[x^2 + c^2 + xy]^2} = 0$

$\Rightarrow c^2 = x(x + y)$

$\Rightarrow \tan\alpha = \frac{cy}{2c^2} = \frac{y}{2c} \Rightarrow y = 2c\tan\alpha$

We had $x^2 + xy - c^2 = 0 \Rightarrow x^2 + 2cx\tan\alpha - c^2 = 0$

Neglecting the negative root we have $\Rightarrow x = -c\tan\alpha + c\sec\alpha$

$\Rightarrow = c\left(\frac{1 - \sin\alpha}{\cos\alpha}\right) = 2d\left(\frac{1 + \tan^2\frac{\alpha}{2} - 2\tan\frac{\alpha}{2}}{1 - \tan^2\frac{\alpha}{2}}\right)$

$= c\left(\frac{1 - \tan\frac{\alpha}{2}}{1 + \tan\frac{\alpha}{2}}\right)$

$= c\tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$

18. The diagram is given below:

We know that $B$ is due north of $D$ at a distance of $2$ km and $D$ is due west of $C$ such that $\angle BCD = 25^\circ$ we can plot $B, C, D$ as shown in the diagram. It is given that $B$ lies on $AC$ such that $\angle BDA = 40^\circ$. From figure it is clear that $\angle ACD = \angle CAD = 25^\circ$ thus $\triangle ACD$ is an isoscelels triangle. Let $AD = CD = x$.

In $\triangle BCD, \tan25^\circ = \frac{2}{x} \Rightarrow x = 2\cot25^\circ = 4.28$ km.

19. The diagram is given below:

Let the train move along the line $PQ$. The train is at $O$ at some instant. $A$ is the observation point. Ten minutes earlier let the train position be $P$ and ten minutes afterwards let the train be at $Q$.

According to question, $\angle OAP = \alpha_1, \angle OAQ = \alpha_2, \angle NOQ = \theta$. Hence $\angle POA = \theta$.

Since the speed of the train is constant $\therefore OP = OQ$

Applying $m:n$ rule in $\triangle PAQ, (1 + 1)\cot\theta = \cot\alpha_2 - \cot\alpha_1$

$\Rightarrow \cot\theta = \frac{\cot\alpha_2 - \cot\alpha_1}{2}$

$\Rightarrow \tan\theta = \frac{2\sin\alpha_1\sin\alpha_2}{\sin(\alpha_1 - \alpha_2)}$

20. The diagram is given below:

Let $OP$ be the flag-staff and that the man walk along the horizontal circle. Clearly, the flag-staff will subtend the greatest and least angles when the man is at $A$ and $B$ respectively. Let $C$ be the mid-point of the arc $ACB$. According to question, $\angle PAO = \alpha, \angle PBO = \beta, \angle PCO = \theta$. Clearly, $\angle POC = 90^\circ$. Let $OP = h, \angle POD = \phi$.

Also, $OA = OB = OC = r$ where $r$ is the radius of the circle.

In $\triangle PDO, \sin\phi = \frac{PD}{OP} \Rightarrow PD = h\sin\phi$

$\cos\phi = \frac{OD}{OP} \Rightarrow OD = h\cos\phi$

$\therefore BD = r + h\cos\phi$ and $AD = r - h\cos\phi$

In $\triangle POC, \tan\theta = \frac{h}{r} \Rightarrow h = r\tan\theta$

In $\triangle PDA, \tan\alpha = \frac{PD}{AD} = \frac{h\sin\phi}{r - h\cos\phi} = \frac{r\tan\theta\sin\phi}{r - r\tan\theta\cos\phi}$

$\Rightarrow \tan\alpha = \frac{\tan\theta\sin\phi}{1 - \tan\theta\cos\phi}$

$\Rightarrow \tan\theta\sin\phi = \tan\alpha - \tan\alpha\tan\theta\cos\phi$

In $\triangle PDB, \tan\beta = \frac{h\sin\phi}{r + h\cos\phi} = \frac{r\tan\theta\sin\phi}{r + r\tan\theta\cos\phi}$

$\Rightarrow \tan\beta = \frac{\tan\theta\sin\phi}{1 + \tan\theta\cos\phi}$

$\Rightarrow \tan\theta\sin\phi = \tan\beta + \tan\beta\tan\theta\cos\phi$

$\Rightarrow \tan\alpha - \tan\alpha\tan\theta\cos\phi = \tan\beta + \tan\beta\tan\theta\cos\phi$

$\Rightarrow \tan\alpha - \tan\beta = \tan\theta\cos\phi(\tan\alpha + \tan\beta)$

Also, $1 - \tan\theta\cos\phi = \frac{\tan\theta\sin\phi}{\tan\alpha}$

and $1 + \tan\theta\cos\phi = \frac{\tan\theta\sin\phi}{\tan\beta}$

$\Rightarrow 2 = \tan\theta\sin\phi\left(\frac{1}{\tan\alpha} + \frac{1}{\tan\beta}\right)$

$\Rightarrow 2\tan\alpha\tan\beta = \tan\theta\sin\phi(\tan\alpha + \tan\beta)$

$\Rightarrow (\tan\alpha - \tan\beta)^2 + 4\tan^2\alpha\tan^2\beta = \tan^2\theta(\tan\alpha + \tan\beta)^2$

$\Rightarrow \tan^2\theta\left[\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}\right]^2 = \left(\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}\right)^2 + \frac{4\sin^2\alpha\sin^2\beta}{\cos^2\alpha\cos^2\beta}$

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

$\Rightarrow \tan\theta = \frac{\sqrt{\sin^2(\alpha - \beta) + 4\sin^2\alpha\sin^2\beta}}{\sin(\alpha + \beta)}$

21. The diagram is given below:

Let $O$ be the position of the observer. $PRQS$ is the horizontal circle in which the bird is flying. $P$ and $Q$ are the two extremes and $R$ is the mid point of the arc of the circle. $P'R'Q'S$ is the vertical projection of the ground. $C$ is the center of the circle $PRQS$.

According to the question, $\angle POP'' = 60^\circ, \angle QOQ' = 30^\circ, \angle ROP'= \theta$.

Also, let $PP' = QQ' = RR' = h, r$ be the radius of the horizontal circle and $OP' = z$.

In $\triangle PP'O, \tan60^\circ = \sqrt{3} = \frac{PP'}{OP'} = \frac{h}{z} \Rightarrow h = \sqrt{3}z$

In $\triangle QOQ', \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{z + 2r} \Rightarrow z + 2r = \sqrt{3}h$

$\Rightarrow z + 2r = \sqrt{3}\sqrt{3}z \Rightarrow z = r$

In $\triangle ROR', \tan\theta = \frac{h}{OR'} = \frac{h}{\sqrt{OC^2 + C'R'^2}} = \frac{h}{\sqrt{(z + r)^2 + r^2}} = \frac{\sqrt{3}r}{\sqrt{(r + r)^2 + r^2}} = \sqrt{\frac{3}{5}}$

$\Rightarrow \tan^2\theta = \frac{3}{5}$

22. The diagram is given below:

Let $O$ be the position of the observer and $OPQ$ be the horizontal line through $O$ meeting the hill at $P$ and the vertical through the center $C$ of the sphere at $Q$.

Let $OA$ be the tangent to the sphere from $O$ touching it at $A$. According to question, $\angle AOQ = \beta, \angle QPC = 90^\circ - \alpha, \angle ACN = \beta$. Let $r$ be the radius of the hill. Draw $AM\perp OQ$ and $AN\perp CR$.

In $\triangle AMO, \tan\beta = \frac{AM}{OM} = \frac{QN}{OP + PM} = \frac{QN}{OP + PQ - MQ}$

$= \frac{CN - CQ}{OP + PQ - MQ}$

$\Rightarrow \frac{\sin\beta}{\cos\beta} = \frac{r\cos\beta - r\cos\alpha}{a + r\sin\alpha - r\sin\beta} [\because MQ = AN]$

$\Rightarrow a\sin\beta + r\sin\beta(\sin\alpha - \sin\beta) = r\cos\beta(\cos\beta - \cos\alpha)$

$\Rightarrow a\sin\beta = r[1 - \cos(\alpha - \beta)]$

$\Rightarrow r = \frac{a\sin\beta}{2\sin^2\frac{\alpha - \beta}{2}}$

Height of the hill above the plane $= OR = CR _ CQ = r - r\cos\alpha = 2r\sin^2\frac{\alpha}{2}$

$= \frac{a\sin\beta\sin^2\frac{\alpha}{2}}{\sin^2\frac{\alpha - \beta}{2}}$

23. The diagram is given below:

Let $O$ be the center of the sphere and $r$ be its radius. Given, $\angle PAM = \theta, \angle PBM = \phi, CA = a, CB = b$

Let $\angle DOC = \beta$

In $\triangle PMA, \tan\theta = \frac{PM}{AM} = \frac{DN}{AC + CM} = \frac{ON _ OD}{AC + DC - DM}$

$\Rightarrow \frac{\sin\theta}{\cos\theta} = \frac{r\cos\theta - r\cos\beta}{a + r\sin\beta - r\sin\theta} [\because DM = NP]$

Proceeding like previous problem $a\sin\beta = r[1 - \cos(\theta - \beta)]$

$\Rightarrow 2r\sin^2\frac{\theta - \beta}{2} = 2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}$

$\Rightarrow \sqrt{r}\sin\frac{\theta - \beta}{2} = \sqrt{a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}$

$\Rightarrow \sqrt{r}\frac{\left[\sin\frac{\theta}{2}\cos\frac{\beta}{2} - \cos\frac{\theta}{2}\sin\frac{\beta}{2}\right]}{\sin\frac{\theta}{2}} = \frac{\sqrt{a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}}{\sin\frac{\theta}{2}}$

$\Rightarrow \sqrt{r}\left[\cos\frac{\beta}{2} - \cot\frac{\theta}{2}\sin\frac{\beta}{2}\right] = \sqrt{a\cot\frac{\theta}{2}}$

Similarly, $\sqrt{r}\left[\cos\frac{\beta}{2} - \cot\frac{\phi}{2}\sin\frac{\beta}{2}\right] = \sqrt{b\cos\frac{\phi}{2}}$

Subtracting, we get $\sqrt{r}\sin\frac{\beta}{2}\left[\cot\frac{\theta}{2} - \cot\frac{\phi}{2}\right] = \sqrt{b\cos\frac{\phi}{2}} - \sqrt{a\cot\frac{\theta}{2}}$

Height of the hill $DR = OR - OD = r - r\cos\beta = 2r\sin^2\frac{\beta}{2}$

$= 2\left[\frac{\sqrt{b\cos\frac{\phi}{2}} - \sqrt{a\cot\frac{\theta}{2}}}{\cot\frac{\theta}{2} - \cot\frac{\phi}{2}}\right]^2$

24. The diagram is given below:

Let $O$ be the center of the hemisphere and $PQ$ is the flag-staff. Given, $OP = OR = r, AB = d$.

In $\triangle ORB, \cos45^\circ = \frac{r}{OB} \Rightarrow OB = \sqrt{2}r$

In $\triangle QOA, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{OQ}{OA} = \frac{h + r}{\sqrt{2}r + d}$

$\Rightarrow h + r = \frac{\sqrt{2}r + d}{\sqrt{3}}$

In $\triangle QOB, \tan45^\circ = 1 = \frac{OQ}{OB} \Rightarrow h + r = \sqrt{2}r$

$\Rightarrow \sqrt{2}r = \frac{\sqrt{2}r + d}{\sqrt{3}}$

$\Rightarrow (\sqrt{6} - \sqrt{2})r = d \Rightarrow r = \frac{\sqrt{3} + 1}{2\sqrt{2}}d$

$\Rightarrow h = (\sqrt{2} - 1)r = \frac{(\sqrt{2} - 1)(\sqrt{3} + 1)}{2\sqrt{2}}d$

25. The diagram is given below:

Let the direction in which man starts walking be the $x$-axis. From question $OA = AB = BC = a$.

Let coordinate of last point be $(X, Y)$ then $X = a + a\cos\alpha + a\cos2\alpha + \cdots$ up to $n$ terms

$= a.\frac{\cos(n - 1)\frac{\alpha}{2}\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}$

$Y = 0 + a\sin\alpha + a\sin2\alpha + \cdots$ up to $n$ terms

$= a\left[\frac{\sin(n - 1)\frac{\alpha}{2}\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}\right]$

Distance from the starting point $= \sqrt{X^2 + Y^2} = \frac{a\sin\frac{n\alpha}{2}}{\sin\frac{\alpha}{2}}$

Let $\theta$ be the angle which this distance makes with $x$-axis. Then

$\tan\theta = \frac{Y}{X} = \tan(n - 1)\frac{\alpha}{2} \Rightarrow \theta = (n - 1)\frac{\alpha}{2}$

26. The diagram is given below:

Let $ABC$ be the horizontal triangle. $A'B'$ represents the stratum of coal. Suppose this startus meets the horizontal plane in line $DD'$. Let $\theta$ be the angle between the horizontal and the stratum of the coal.

Clearly, $\angle ADA' = \theta$. According to question, $AA' = x, BB' = x + y$ and $CC' = x + z$.

In $\triangle AA'D, \tan\theta = \frac{AA'}{AD} \Rightarrow x = AD\tan\theta$

In $\triangle BB'D, \tan\theta = \frac{BB'}{BD} \Rightarrow x + y = (AD + AB)\tan\theta$

$\Rightarrow x + z = (AD + c)\tan\theta$

In $\triangle CC'D, \tan\theta = \frac{CC'}{CD'} \Rightarrow x + z = (AD + b\cos A)\tan]\theta$

$\Rightarrow y = c\tan\theta \Rightarrow \frac{y}{c} = \tan\theta$ and $z = b\cos A\tan\theta \Rightarrow \frac{z}{b} = \cos A\tan\theta$

Now, $\frac{y^2}{c^2} + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A = \frac{y^2}{c_2}\sin^2A + \frac{y^2}{c^2}\cos^2A + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A$

$= \frac{y^2}{c^2}\sin^2A + \left(\frac{y}{c}\cos A - \frac{z}{b}\right)^2$

$= \tan^2\theta\sin^2A + (\tan\theta\cos A - \cos A\tan\theta)^2 = \tan^2\theta\sin^2A$

$\Rightarrow \tan\theta\sin A = \sqrt{\frac{y^2}{c^2} + \frac{z^2}{b^2} - \frac{2yz}{bc}\cos A}$