# 30. Height and Distance Solutions Part 3#

1. The diagram is given below:

Let $AB$ and $CD$ be the two temples and $AC$ be the river. Let the height of temple $AB$ be $50$ m. $AC$ is the river. The angles are depression are shown as corresponding angles of elevation. Let the height of $CD$ be $x$ m and width of river be $w$ m. Thus, $CD = x$ and $AC = w$.

In $\triangle ABC, \tan60^\circ = \frac{AB}{AC} \Rightarrow w = \frac{50}{\sqrt{3}} = DE[\because DE||AC]$

In $\triangle BDE, \tan30^\circ = \frac{BD}{DE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BD}{50/\sqrt{3}}\Rightarrow BD = \frac{50}{3}$.

Thus, height of the second temple $CD = AB - BD = \frac{100}{3}$ m.

2. This problem is similar to $95$ and has been left as an exercise.

3. This problem is similar to $95$ and has been left as an exercise.

4. The diagram is given below:

Let $BE$ be the tower leaning due east where $B$ is the foot of the tower and $E$ is the top. $AB$ is the vertical height of the tower taken as $h$. The angles of elevation are shown from tow points as given in the question.

In $\triangle ACE, \tan\alpha = \frac{h}{x + a}$ and in $\triangle ADE, \tan\beta = \frac{h}{x + b}$.

$\Rightarrow \frac{b - a}{h} = \frac{1}{\tan\beta} - \frac{1}{\tan\alpha}$

$\Rightarrow h = \frac{(b - a)\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}$.

5. The diagram is given below:

Let $AC$ be the lake and $B$ be the point of observation $2500$ m above lake. Let $E$ be the cloud and $F$ be its reflection in the lake. If we take height of the cloud above lake as $h$ then $CD = 2500$ m where $BD||AC$ . $DE = h - 2500$ m and $CD = 2500$ m. The angle of elevation and angle of depression of cloud and its reflection are shown as given in the problem.

In $\triangle BDF, \tan45^\circ = \frac{DF}{BD} \Rightarrow BD = h + 2500$

In $\triangle BDE, \tan15^\circ = \frac{DE}{BD} \Rightarrow DE = 1830.6$ m.

$\Rightarrow CE = CD + DE = h = 2500 + 1830.6 = 4330.6$ m.

6. The diagram is given below:

This is a problem similar to previous problem with $2500$ replaced by $h$ and angles are replaced by $\alpha$ and $\beta$. So the diagram is similar in nature. Let the height of the cloud above lake be $h'$ m. So $DE = h' - h$ and $DF = h + h'$.

In $\triangle BDE, \tan\alpha = \frac{DE}{BD}\Rightarrow BD = (h' - h)/\tan\alpha$

In $\triangle BDF, \tan\beta = \frac{DF}{BD}\Rightarrow BD = (h' + h)/\tan\beta$

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

$\therefore BD = \frac{2h}{(\tan\alpha - \tan\beta)}$

Also, $\sec\alpha = \frac{BE}{BD} \Rightarrow BE = \frac{2h\sec\alpha}{\tan\alpha - \tan\beta}$ which is the distance of the cloud from the point of observation.

7. The diagram is given below:

Let $AB$ be the height of plane above horizontal ground as $h$ miles. $C$ and $D$ are two consecutive milestones so $CD = 1$ mile. Let $BC = x$ mile. The angles of depression are represented as angles of elevation.

In $\triangle ABC, \tan\alpha = \frac{AB}{BC} \Rightarrow h = x\tan\alpha$

In $\triangle ABD, \tan\beta = \frac{AB}{BD} \Rightarrow h = (x + 1)\tan\beta$

$\Rightarrow x = \frac{\tan\beta}{\tan\alpha - \tan\beta}\Rightarrow h = \frac{\tan\alpha}{\tan\alpha - \tan\beta}$.

8. The diagram is given below:

Let $PQ$ be the post with height $h$ and $AB$ be the tower. Given that the angles of elevation of $B$ at $P$ and $Q$ are $\alpha$ and $\beta$ respectively. Draw $CQ||PA$ such that $PQ = AC = h$ and A:math:AP = QC = x. Also, let $BC = h'$ so that $AB = AC + BC = h + h'$.

In $\triangle ABP, \tan\alpha = \frac{h + h'}{x}\Rightarrow x = \frac{h + h'}{\tan\alpha}$

In $\triangle BCQ, \tan\beta = \frac{h'}{x}\Rightarrow x = \frac{h'}{\tan\beta}$

$\Rightarrow \frac{h + h'}{\tan\alpha} = \frac{h'}{\tan\beta}$

$h' = \frac{h\tan\beta}{\tan\alpha - \tan\beta} \Rightarrow x = \frac{h}{\tan\alpha - \tan\beta}$

$\therefore PQ = h + h' = \frac{h\tan\alpha}{\tan\alpha - \tan\beta}$.

9. The diagram is given below:

Let $AD$ be the wall, $BD$ and $CE$ are two positions of the ladder. Then according to question $BC = a, DE = b$ and angles of elevations at $B$ and $C$ are $\alpha$ and $\beta$. Let $AB = x$ and $AE = y$. Also, let length of ladder be $l$ i.e $BD = CE = l$.

In $\triangle ABD, \sin\alpha = \frac{AD}{BD} = \frac{y + b}{l}, \cos\alpha = \frac{AB}{BD} = \frac{x}{l}$

In $\triangle ACE, \sin\alpha = \frac{AE}{CE} = \frac{y}{l}, \cos\beta = \frac{AC}{CE} = \frac{a + x}{l}$.

$\Rightarrow \cos\beta - \cos\alpha = \frac{a}{l}$ and $\sin\alpha - \sin\beta = \frac{b}{l}$

$\Rightarrow \frac{a}{b} = \frac{\cos\alpha - \cos\beta}{\sin\beta - \sin\alpha}$.

10. The diagram is given below:

Let $CD$ be the tower subtending angle $\alpha$ at $A$. Let $B$ be the point $b$ m above $A$ from which angle of depression to foot of tower at $C$ is $\beta$ which is shown as angle of elevation. Let $AC = x$ and $CD = h$.

In $\triangle ACD, \tan\alpha = \frac{h}{x} \Rightarrow x = h\cot\alpha$

In $\triangle ABC, \tan\beta = \frac{b}{x} \Rightarrow x = b\cot\beta$

$\Rightarrow h\cot\alpha = b\cot\beta \Rightarrow h = b\tan\alpha\cot\beta$

11. The diagram is given below:

Let $AB$ be the observer with a height of $1.5$ m, $28.5$ m i.e. $AD$ from tower $DE, 30$ m high. Draw $BC||AD$ such that $AB = CD = 1.5$ m and thus $CE = 28.5$ m. Let the angle of elevation from observer’s eye to the top of the tower be $\alpha$.

In $\triangle BCE, \tan\alpha = \frac{CE}{BC} = \frac{28.5}{28.5} = 1\Rightarrow \alpha = 45^\circ$.

12. The diagram is given below:

Let $AB$ be the tower havin a height of $h$ and $C$ and $D$ are two objects at a distance of $x$ and $x + y$ such that angles of depression shown as angles of elevatin are $\beta$ and $\alpha$ respectively.

In $\triangle ABC, \tan\beta = \frac{h}{x} \Rightarrow x = h\cot\beta$

In $\triangle ABD, \tan\alpha = \frac{h}{x + y} \Rightarrow x + y = h\cot\alpha$

Distance between $C$ and $D = y = h(\cot\alpha - \cot\beta)$.

13. The diagram is given below:

Let $AB$ be the height of the window at a height $h$ and $DE$ be the house opposite to it. Let the distance between the houses be $AD = x$. Draw $BC||AD$ such that $BC = x$ and $CD = h$. The angles are shown as given in the problem. Let $CE = y$

In $\triangle BCD, \tan\beta = \frac{h}{x} \Rightarrow x = h\cot\beta$

In $\triangle BCE, \tan\alpha = \frac{y}{x} \Rightarrow x = y\cot\alpha$

$\Rightarrow y = h\tan\alpha\cot\beta$

Total height of the second house $DE = CD + DE = y + h = h(1 + \tan\alpha\cot\beta)$

14. The diagram is given below:

Let $AD$ be the ground, $B$ be the lower window at a height of $2$ m, $C$ be the upper window at a height of $4$ m above lower window and $G$ be the balloon at a height of $x + 2 + 4$ m above ground. Draw $DG||AC, BE||AD$ and $CF||AD$ so that $DE = 2$ m, $EF = 4$ m and $FG = x$ m. Also, let $BE = CF = d$ m. The angles of elevation are shown as given in the problem.

In $\triangle BEG, \tan60^\ circ = \sqrt{3} = \frac{EG}{BE} = \frac{x + 4}{d}\Rightarrow d = \frac{x + 4}{\sqrt{3}}$

In $\triangle CFG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{FG}{CF} = \frac{x}{d}\Rightarrow d = \sqrt{3}x$

$\Rightarrow \sqrt{3}x = \frac{x + 4}{\sqrt{3}}\Rightarrow x = 2$

$\therefore$ the height of the balloon $= 2 + 4 + 2 = 8$ m.

15. The diagram is given below:

Let $AB$ be the lamp post, $EF$ and $GH$ be the two positions of the man having height $6$ ft. Let the shdows be $EC$ and $GD$ of lengths $24$ ft. and $30$ ft. for initial and final position. Since the man moves eastward from his initial position $\therefore \angle ACD = 90^\circ$.

Let $AB = h, AE = x$ and $AG = y$.

From similar triangles $CEF$ and $ABC$

$\frac{h}{6} = \frac{24 + x}{24}$

From similar triangles $ABD$ and $DGH$

$\frac{h}{6} = \frac{30 + y}{30}$

Thus, $1 + \frac{x}{24} = 1 + \frac{y}{30} \Rightarrow y = \frac{5x}{4}$.

From right angles $\triangle ACD$,

$y^2 = x^2 + 300^2 \Rightarrow x = 400$ ft.

$\Rightarrow h = 106$ ft.

16. The diagram is given below:

Let $AB$ be the tower having a height of $h$ m, $AC$ be the final length of shadow taken as $x$ m, $AD$ is the initial length of shadow which is $5$ m more than finla length i.e. $CD = 5$ m. The angles of elevation are shown as given in the problem.

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

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 5} \Rightarrow \frac{2h}{\sqrt{3}} = 5$

$\Rightarrow h = \frac{5\sqrt{3}}{2} = 4.33$ m.

17. The diagram is given below:

Let $A$ be the initial position of the man and $D$ and $E$ be the objects in the west. Let $DE = x, AD = y, \angle ADB=\theta, \angle AEB = \phi$ and $\angle ADC=\psi$. $\alpha$ and $\beta$ are the angles made by objects on the two positions of the man as given in the problem.

$\Rightarrow \tan\theta = \frac{c}{y}$ and $\tan\phi = \frac{c}{x + y}$

Now $\theta - \phi = \alpha \Rightarrow \tan(\theta - \phi) = \tan\alpha$

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

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

$\Rightarrow cx\cot\alpha = xy + y^2 + c^2$

Similarly, substituting $2c$ for $x$ and $\psi$ for $\phi$, we get

$2cx\cot\beta = xy + y^2 + 4c^2 \Rightarrow x= \frac{3c}{2\cot\beta - \cot\alpha}$.

18. The diagram is given below:

Let $P$ be the object and $OA$ be the straight line on which $B$ and $C$ lie underneath the object. Let $OP = h$. According to question the angles of elevation made are $\alpha, 2\alpha$ and $3\alpha$ from $A, B$ and $C$ i.e. $\angle PCO = 3\alpha, \angle PBO = 2\alpha$ and $\angle PAO = \alpha$. Given that $AB = \alpha$ and $BC = b$.

$\angle APB = 2\alpha - \alpha = \alpha$ and $\angle BPC = 3\alpha - 2\alpha = \alpha$

$\therefore AB = BP = a$

In $\triangle PBC, \frac{BC}{\sin\alpha} = \frac{PB}{\sin(180^\circ - 3\alpha)} \Rightarrow \frac{b}{\sin\alpha} = \frac{a}{\ sin3\alpha}$

$\Rightarrow \frac{a}{b} = \frac{\sin3\alpha}{\sin\alpha} = 3 - 4\sin^2\alpha \Rightarrow \sin\alpha = \sqrt{\frac{3b - a}{4b}}$

In $\triangle OPB, OP = BP\sin2\alpha = 2a\sin\alpha\cos\alpha = \frac{a}{2b}\sqrt{(a + b)(3b - a)}$

19. This problem is similar to $92$ and has been left as an exercise.

20. The diagram is given below:

Let $\theta$ be the angle of inclination of the inclines plane $AC$. Let $AB = c$ and $BC = c$. Let the object be at $D$. Now $\angle DBA = \theta - \alpha$ and $\angle DCA = \theta - \beta$.

Using sine rule in $\triangle DAB, \frac{c}{\sin\alpha} = \frac{AD}{\sin(\theta - \alpha)}$

$\Rightarrow AD = \frac{c\sin(\theta - \alpha)}{\sin\alpha}$

Applying sine rule in $\triangle DAC, \frac{2c}{\sin\beta} = \frac{AD}{\sin(\theta - \beta)}$

$\Rightarrow AD = \frac{2\sin(\theta - \beta)}{\sin\beta}$

$\Rightarrow \frac{c\sin(\theta - \alpha)}{\sin\alpha} = \frac{2c\sin(\beta - \beta)}{\sin\beta}$

$\Rightarrow \frac{\sin\theta\cos\alpha - \cos\theta\sin\alpha}{\sin\theta\sin\alpha} = \frac{2[\sin\theta\cos\beta - \cos\theta\sin\beta]}{\sin\theta\sin\beta}$

$\Rightarrow \cot\alpha - \cot\theta = 2(\cot\beta - \cot\beta)$

$\Rightarrow \cot\theta = 2\cot\beta - \cot\alpha$.

21. The question is same as $109$ just that we have a different relation to prove. From $109$ we have

$\frac{a}{b} = \frac{\cos\beta - \cos\alpha}{\sin\alpha - \sin\beta}$

$= \frac{2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}{2\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}}$

$= \tan\frac{\alpha + \beta}{2}\Rightarrow a = b\tan\frac{\alpha + \beta}{2}$.

22. The diagram is given below:

Given that $A$ and $B$ are two points of observation on ground $1000$ m apart. Let $C$ be the point where the balloon will hit the ground at a distance $x$ m from $B$. Also, let $D$ and $E$ be the points above $A$ and $B$ respectively such that $\angle BAE= 30^\circ$ and $\angle DBA = 60^\circ$.

In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{AD}{AB}\Rightarrow AD = 1000\sqrt{3}$ m.

In $\triangle ABE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{AB}\Rightarrow BE = \frac{1000}{\sqrt{3}}$ m.

Clearly, $\triangle BCE$ and $ACD$ are similar. Therefore,

$\frac{BC}{AC} = \frac{BE}{AD} \Rightarrow \frac{x}{x + 1000} = \frac{1000}{1000\sqrt{3}.\sqrt{3}}$

$\Rightarrow x = 500 \Rightarrow AC = 1500$ m.

23. The diagram is given below:

Let $AB$ be tree having height $h$ m and $BC$ be the width of the river having width $w$ m. According to question angle of elevation of the tree from the opposite bank is $60^\circ$. Also, let $D$ be the point when the man retires $40$ m from where the angle of elevation of the tree is $30^\circ$.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{h}{w} \Rightarrow h = w\sqrt{3}$ m.

In $\triangle ABD, \tan30^\circ = \frac{h}{w + 40}\Rightarrow 3w = w + 40 \Rightarrow w = 20$ m.

$\Rightarrow h = 20\sqrt{3}$ m.

Thus, width of the river is $20$ m and height of the tree is $20\sqrt{3}$ m.

24. The diagram is given below:

Let $O$ be the point of observation. The bird is flying in the horizontal line $WXYZ$. The angles of elevation of the bird is given at equal intervals of time. Since the speed of the bird is constant $WX = XY = YZ = y$ (let). From question $\angle AOW = \alpha, \angle BOX = \beta, \angle COY = \gamma$ and $\angle DOZ = \delta$. Let $OA = x$ and $AW = h$.

In $\triangle AOW, \cot\alpha = \frac{x}{h}$

In $\triangle BOX, \cot\beta = \frac{x + y}{h}$

In $\triangle COY, \cot\gamma = \frac{x + 2y}{h}$

In $\triangle DOZ, \cot\delta = \frac{x + 3y}{h}$

L.H.S. $= \cot^2\alpha - \cot^2\delta = \frac{-6xy - 9y^2}{h^2}$

R.H.S. $= \cot^2\beta - \cot^2\gamma = \frac{-6xy - 9y^2}{h^2}$

$\therefore$ L.H.S. = R.H.S.

25. The diagram is given below:

Let $AB$ be the tower, $BC$ be the pole and $D$ be the point of observation where the tower and the pole make angles $\alpha$ and $\beta$ respectively. Let the height of the tower be $h'$ and $AD = d$. Given that the height of the pole is $h$.

In $\triangle ABD, \cot\alpha = \frac{AD}{AB} = \frac{d}{h'} \Rightarrow d = h'\cot\alpha$

In $\triangle ACD, \tan(\alpha + \beta) = \frac{AC}{AB} = \frac{h + h'}{d}$

$\Rightarrow h + h' = h'\cot\alpha\tan(\alpha + \beta)$

$h' = \frac{h}{\cot\alpha\tan(\alpha + \beta) - 1} = \frac{h\sin\alpha\cos(\alpha + \beta)}{\sin(\alpha + \beta)\cos\alpha - \cos(\alpha + \beta)\sin\alpha}$

$= \frac{h\sin\alpha\cos(\alpha + beta)}{\sin(\alpha + \beta - \alpha)} = h\sin\alpha\cosec\beta\cos(\alpha + \beta)$

26. The diagram is given below:

Given $AC = BC = x$ (let) and $\angle BPC = \beta$.

Let $\angle BPA = \theta$ then $\angle CPA = \theta - \beta$

In $\triangle APC, \tan(\theta - \beta) = \frac{x}{AP}$

In $\triangle APB, \tan\theta = \frac{2x}{AP}$

$\Rightarrow \tan\theta = 2\tan(\theta - \beta) = \frac{2(\tan\theta - \tan\beta)}{1 + \tan\theta\tan\beta}$

$\tan\theta = \frac{AB}{AP} = \frac{1}{n}$ (from question)

$\Rightarrow \frac{1}{n} = \frac{2\left(\frac{1}{n} - \tan\beta\right)}{1 + \frac{\tan\beta}{n}}$

$\Rightarrow \tan\beta = \frac{n}{2n^2 + 1}$.

27. The diagram is given below:

Let $AB$ be the first chimney and $CD$ be the second chimney. The angles of elevation are shown as angles of elevation as given in the problem. Draw $BE||AC$ and let $AC = BE = d$ m and $AB = CE = h$ m. Given $CD = 150$ m. Clearly, $DE = 150 - h$ m.

In $\triangle BED, \tan\theta = \frac{4}{3} = \frac{150 - h}{d} \Rightarrow 4d = 450 - 3h$

In $\triangle ACD, \tan\phi = \frac{5}{2} = \frac{150}{d} \Rightarrow d = 60$ m.

$\Rightarrow h = 70$ m. $\Rightarrow BE = 60$ m and $ED = 150 - 70 = 80$ m.

$BD^2 = BE^2 + DE^2 = 80^2 + 60^2 \Rightarrow BD = 100$ m, which, is the distance between the tops of two chimneys.

28. The diagram is given below:

Let $CD$ be the tower of height $h$ having an elevation of $30^\circ$ from $A$ which is southward of it. Let $B$ be eastward of $A$ at a distance of $a$ from it from where the angle of elevation is $18^\circ$. Since $B$ is eastward of $A \angle CAB = 90^\circ$.

In $\triangle ACD, \tan 30^\circ = \frac{h}{AC} \Rightarrow AC = h\sqrt{3}$

In $\triangle BCD,, \tan18^\circ = \frac{h}{BC} \Rightarrow BC = h\cot18^\circ$

In $\triangle ABC, BC^2 = a^2 + AC^2 \Rightarrow h^2\cot^218^\circ = a^2 + 3h^2$

$\therefore h = \frac{a}{\sqrt{\cot^218^\circ - 3}}$

Now $\cot^218^\circ = 5 + 2\sqrt{5} \therefore h = \frac{a}{\sqrt{2 + 2\sqrt{5}}}$.

29. The diagram is given below:

Let $AB$ be the tower having height $h$. Given that $P$ is north of the tower and $Q$ is due west of $P\therefore \angle APQ= 90^\circ$.

In $\triangle ABP, \tan\theta = \frac{h}{AP} \Rightarrow AP = h\cot\theta$

In $\triangle ABQ, \tan\phi = \frac{h}{AQ} \Rightarrow AQ = h\cot\phi$

In $\triangle APQ, AQ^2 = AP^2 + PQ^2$

$\Rightarrow PQ^2 = h^2[\cot^2\phi - \cot^2\theta]$

$\Rightarrow h = \frac{PQ}{\sqrt{\cot^2\phi - \cot^2\theta}}$

$= \frac{PQ\sin\theta\sin\phi}{\sqrt{\sin^2\theta\cos^2\phi - \sin^2\phi\cos^2\theta}}$

$= \frac{PQ\sin\theta\sin\phi}{\sqrt{(\sin^2\theta(1 - \sin^2\phi)) - \sin^2\phi(1 - \sin^2\theta)}}$

$= \frac{PQ\sin\theta\sin\phi}{\sqrt{\sin^2\theta - \sin^2\phi}}$.

30. The diagram is given below:

Let $B$ be the peak having a height of $h$ with base $A$. Let $PQ$ is the horizontal base having a length $2a$ making angle of elevation of $\theta$ from each end. Let $R$ be the mid-point of $PQ$ from where the angle of elevation of $B$ is $\phi$ as given in the question.

Thus, $\angle APB = \angle AQB = \theta$ and $\angle ARB = \phi$.

In $\triangle APB, \tan\theta = \frac{h}{AP} \Rightarrow AP = h\cot\theta$

Similarly, $AQ = h\cot\theta$ and $AR = h\cot\phi$

$\because AR$ is the median of the $\triangle APQ$

$\therefore AP^2 + AQ^2 = 2PR^2 + 2AR^2$

$\Rightarrow 2h^2\cot^2\theta = 2a^2 + 2h^2\cot^2\phi$

$\Rightarrow h^(\cot^2\theta - \cot^2\phi) = a^2$

$\Rightarrow h = \frac{a}{\sqrt{\frac{\cos^2\theta}{\sin^2\theta} - \frac{\cos^2\phi}{\sin^2\phi}}}$

$= \frac{a\sin\theta\sin\phi}{\sqrt{(\sin\phi\cos\theta + \cos\phi\sin\theta)(\sin\phi\cos\theta - \cos\phi\sin\theta)}}$

$= \frac{a\sin\theta\sin\phi}{\sqrt{\sin(\theta + \phi)\sin(\phi - \theta)}}$

31. The diagram is given below:

Let $B$ be the top of the hill such that height of the hill $AB$ is $h$ and $P, R, Q$ be the three consecutive milestones. Given, $\angle APB = \alpha, \angle ARB = \beta, \angle AQB = \gamma$.

In $\triangle APB, \tan\alpha = \frac{h}{AP}\Rightarrow AP = h\cot\alpha$

Similalrly, $AR = h\cot\beta$ and $AQ = h\cot\gamma$

Also, $PR = QR = 1$ mile.

$\because PR = QR, AR$ is the median of the triangle $APR$.

$\therefore AP^2 + AR^2 = 2PR^2 + 2AQ^2 \Rightarrow h^2(\cot^2\alpha + \cot^2\gamma) = 2 + 2h^2\cot^2\beta$

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

32. The diagram is given below:

Let $OP$ be the tower haing a height of $h$ which is to be found. Let $ABC$ be the equilateral triangle. Given that $OP$ subtends angles of $\alpha, \beta, \gamma$ at $A, B, C$ respectively. Given that $\tan\alpha = \sqrt{3} + 1$ and $\tan\beta = \tan\gamma = \sqrt{2}$. It is given that $OP$ is perpendicular to the plane of $\triangle ABC$.

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

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

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

$\therefore \angle OAB = \angle OAC$ but $\angle BAC = 60^\circ$

$\Rightarrow \angle OAB = \angle OAC = 30^\circ$

Using sine rule in the $\triangle OAB, \frac{OB}{\sin30^\circ} = \frac{OA}{\sin\theta}$ (let $\angle ABO = \theta$)

$\Rightarrow \frac{\frac{h}{\sqrt{2}}}{\frac{1}{2}} = \frac{\frac{h}{\sqrt{3} + 1}}{\sin\theta}$

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

$\Rightarrow \theta = 15^\circ$

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

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

In $\triangle OBD, \cos45^\circ = \frac{BD}{OB} \Rightarrow \frac{1}{\sqrt{2}} = \frac{40}{h\sqrt{2}}\Rightarrow h = 80'$

33. The diagram is given below:

In the diagram we have shown only one tower instead of three. We will apply cyclic formula to this one tower relationships. Let $P$ be the position of the eye and height of $PQ = x$. Let $AB$ be the tower having a height of $a$ as given in the question and let the angle subtended by $AB$ at $P$ is $\theta$.

Thus, $\angle APB = \theta, \angle PAQ = \alpha \Rightarrow \angle ABP = 180^\circ - \theta - (90^\circ - \alpha) = 90^\circ + (\alpha - \theta)$

By sine rule in $\triangle APB$,

$\frac{a}{\sin\theta} = \frac{AP}{\sin[90^\circ + (\alpha - \theta)]} = \frac{AP}{\cos(\alpha - \theta)}$

In $\triangle APQ, \sin\alpha = \frac{x}{AP} \Rightarrow AP = \frac{x}{\sin\alpha}$

$\Rightarrow \frac{a}{\sin\theta} = \frac{x}{\sin\alpha\cos(\alpha - \theta)}$

$\Rightarrow x\sin\theta = a\sin\alpha\cos(\alpha - \theta)\Rightarrow \cos(\alpha - \theta) = \frac{x\sin\theta}{a\sin\alpha}$

If we consider the other two towers we will have similar relations i.e.

$\Rightarrow \cos(\beta - \theta) = \frac{x\sin\theta}{b\sin\beta}$ and $\cos(\gamma - \theta) = \frac{x\sin\theta}{c\sin\gamma}$

Now, $\frac{\sin(\beta - \gamma)}{a\sin\alpha} + \frac{\sin(\gamma - \alpha)}{b\sin\beta} + \frac{\sin(\alpha - \beta)}{c\sin\gamma}$

$=\displaystyle\sum\frac{\sin(\alpha - \beta)}{c\sin\gamma} = \displaystyle\sum\frac{\sin(\alpha - \theta + \theta - \beta)}{c\sin\gamma}$

$= \displaystyle\sum\frac{\sin(\alpha - \theta)\cos(\theta - \beta) + \cos(\alpha - \theta)\sin(\theta - \beta)}{c\sin\gamma}$

$=\displaystyle\sum\frac{1}{c\sin\gamma}\left[\frac{\sin(\alpha - \theta)x\sin\theta}{b\sin\beta} + \frac{\sin(\theta - beta)x\sin\theta}{a\sin\alpha}\right]$

$= \displaystyle\sum\frac{x\sin\theta}{abc\sin\alpha\sin\beta\sin\gamma}\left[a\sin(\alpha - \theta)\sin\alpha - b\sin(\beta - \theta)sin\beta\right]$

$= \frac{x\sin\theta}{abc\sin\alpha\sin\beta\sin\gamma}.0 = 0$

34. The diagram is given below:

Let $S$ be the initial position of the man and $P$ and $Q$ be the poosition of the objects. Since $PQ$ subtends greatest angle at $R$, a circle will pass through $P, Q$ and $R$ and $RS$ will be a tangent to this circle at $R$.

Also, $\angle PQR = \angle PRS = \theta$ (let). Let $PQ = x$.

Clearly $\angle SRQ = \theta + \beta$

Using sine law in $\triangle PRQ, \frac{x}{\sin\beta} = \frac{PR}{\sin\theta} \Rightarrow x = \frac{PR\sin\beta}{\sin\theta}$

Using sine law in $\triangle PRS, \frac{PR}{\sin\alpha} = \frac{c}{\sin(\theta + \beta)} \Rightarrow PR = \frac{c\sin\alpha}{\sin(\theta + \beta)}$

$\Rightarrow x = \frac{c\sin\alpha\sin\beta}{\sin(\theta + \beta)\sin(\theta)} = \frac{2c\sin\alpha\sin\beta}{2\sin(\theta + \beta)\sin(\theta)}$

$= \frac{2c\sin\alpha\sin\beta}{\cos\beta - \cos(2\theta + \beta)}$

In $\triangle QRS, \alpha + \beta + 2\theta = 180^\circ \Rightarrow 2\theta + \beta = 180^\circ - \alpha$

$\Rightarrow \cos(2\theta + \beta) = -\cos\alpha$

$\Rightarrow x = \frac{2c\sin\alpha\sin\beta}{\cos\alpha + \cos\beta}$.

35. 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 $x$. $A$ and $B$ are the two points on the horizontal line $OA$. Let $OB = y$. Given, $AB = d, \angle QAP = \angle QBP = \alpha$.

Since $\angle QAP = \angle QBP$, a circle will pass through the points $A, B, P$ and $Q$ because angles in the same segment of a circle are equal.

Thus, $\angle BAP = \angle BQP = \beta$ (angles on the same segment $BP$)

$\Rightarrow \angle BPO = \angle QAO = \alpha + \beta$

In $\triangle AOP, \tan\beta = \frac{h}{y + d}$

In $\triangle BOP, \tan(\alpha + \beta) = \frac{y}{h} \Rightarrow y = h\tan(\alpha + \beta)$

$\Rightarrow h = y\tan\beta + d\tan\beta\Rightarrow h = \frac{d\tan\beta}{1 - \tan(\alpha + \beta)\tan\beta}$

In $\triangle BOQ, \tan\beta = \frac{y}{x + h}\Rightarrow x\tan\beta + h\tan\beta = h\tan(\alpha + \beta)$

$\Rightarrow x = \frac{d[\tan(\alpha + \beta) - \tan\beta]}{1 - \tan(\alpha + \beta) + \tan\beta}$

36. This question is same as $92$ with $\alpha$ replaced by $\theta$ and $\beta$ replaced by $\phi$.

Referring to diagram of $92, AC = \frac{h(\tan\theta + \tan\phi)}{\tan\phi - \tan\theta}$

$= \frac{h(\sin\theta\cos\phi + \sin\phi\cos\theta)}{\sin\phi\cos\theta - \sin\theta\cos\phi}$

$= \frac{h\sin(\theta + \phi)}{\sin(\phi - \theta)}$.

37. The diagram is given below:

Let $BC$ represent the road inclined at $10^\circ$ to the vertical towards sun and $AB = 2.05$ m represents the shadow where the elevation of the sun is $\angle BAC = 38^\circ$. Thus, $\angle BCA = 180^\circ - (10^\circ + 90^\circ + 38^\circ) = 42^\circ$.

Using sine rule in $\triangle ABC$,

$\frac{BC}{\sin38^\circ} = \frac{AB}{\sin426\circ}\Rightarrow BC = \frac{2.05\sin38^\circ}{\sin42^\circ}$.

38. The diagram is given below:

Let $CD$ be the tower having a height of $h$ m. Let $BC$ be its shadow when altitude of the sun is $60^\circ$ and $AC$ be its shadow when altitude of the sun is $30^\circ$.

Given that shadow decreases by $30$ m when altitude changes from $30^\circ$ to $60^\circ$ i.e. $AB = 30$ m. Let $BC = x$ m.

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

In $\triangle ACD, \tan30^\circ = \frac{h}{x + 30} \Rightarrow h = 15\sqrt{3}$ m.

39. This problem is similar to $138$ and has been left as an exercise.

40. This problem is similar to $96$ and has been left as an exercise. The answer is $90$ seconds.

41. The diagram is given below:

Let $C$ be the position of the aeroplane flying $3000$ m above ground and $D$ be the aeroplane below it. Given that the angles of elevation of these aeroplanes are $45^\circ$ and $60^\circ$ respectively. Let the height of $D$ is $h$ m and $AB = d$ m.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{3000}{d} \Rightarrow d = 1000\sqrt{3}$ m.

In $\triangle ABD, \tan45^\circ = 1 = \frac{h}{d} \Rightarrow h = 1000\sqrt{3}$ m.

$\therefore$ Distance between heights of the aeroplanes $= CD = 3000 - 1000\sqrt{3} = 1268$ m.

42. The diagram is given below:

Let $C$ and $D$ be two consecutive milestones so that $CD = 1$ mile. Let $D$ be position of aeroplane having a height $h$ above $A$, to which angles of elevation are $\alpha$ and $\beta$ from $C$ and $D$ respectively. Let $AC = x \Rightarrow AD = 1 - x$.

In $\triangle ABC, \tan\alpha = \frac{h}{x} \Rightarrow h = x\tan\alpha$

In $\triangle ABD, \tan\beta = \frac{h}{1 - x} \Rightarrow h = \frac{\tan\alpha\tan\beta}{\tan\alpha + \tan\beta}$

43. This problem is similar to $119$ and has been left as an exercise.

44. The diagram is given below:

Using $m:n$ theorem from section 16.9,

$2c\cot(\theta - 30^\circ) = c\cot15^\circ - c\cot30^\circ$

$\Rightarrow \cot(\theta - 30^\circ) = 1 = \cot45^\circ$

$\Rightarrow \theta = 75^\circ$.

45. This problem is simmilar to $138$, and has been left as an exercise.

46. The diagram is given below:

Let $AB$ be the height of air-pilot which has height of $h$. Let $CD$ be the tower whose angles of depression of top and bottom of tower be $30^\circ$ and $60^\circ$ respectively. Draw $DE||AC$ such that $DE = AC$. Let the height of tower $CD$ be $x$.

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

In $\triangle ADE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{x}\Rightarrow BE = \frac{x}{\sqrt{3}} = \frac{h}{3}$

$\therefore$ Height of the tower $CD = h - \frac{h}{3} = \frac{2h}{3}$