Question 1: Find $AD$:

$\frac{AE}{BE}$ $= \tan 32^o$

$\Rightarrow AE = 20 \times \tan 32^o = 12.$

$AD = 5 + 12.5 = 17.5 \ m$

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Question 2: In the following diagram, $AB$ is a floor-board; $PQRS$ is a cubical box with each edge $= 1 \ m$ and $\angle B = 60^o$. Calculate the length of the board $AB$.

$\frac{PS}{PB}$ $= \sin 60^o$

$PB =$ $\frac{1}{\sin 60^o}$ $= 1.15$

$\frac{PQ}{AP}$ $= \cos 60^o$

$AP =$ $\frac{1}{\cos 60^o}$ $= 2$

$\therefore AB = PB + AP = 1.15 + 3 = 3.15 \ m$

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Question 3: Calculate $BC$.

$\frac{CD}{20}$ $= \tan 42^o$

$CD = 20 \tan 42^o = 18$

$\frac{20}{BC+18}$ $= \tan 35^o$

$20 = BC \tan 35^o + 18 \tan 35^o$

$BC =$ $\frac{20 - 18 \tan 35^o}{\tan 35^o}$ $= 10.56$

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Question 4: Calculate $AB$.

$\frac{AX}{6}$ $= \cos 30^o$

$AX = 6 \cos 30^o = 5.196 \ m$

$\frac{XB}{5}$ $= \cos 43^o$

$XB = 5 \cos 43^o = 3.856 \ m$

$AB = 5.196 + 3.856 = 8.85 \ m$

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Question 5: The radius of a circle is given as $15 \ cm$ and chord $AB$ subtends an angle of $131^o$ at the center $C$ of the circle. Using trigonometry, calculate: (i) the length of $AB$; (ii) the distance of $AB$ from the center $C$.

$\frac{AX}{15}$ $= \cos 24.5^o$

$AX = 15 \cos 24.5^o = 13.64 \ cm$

Therefore $AB = 2 \times 13.64 = 27.3 \ cm$

$\frac{OX}{15}$ $= \sin 24.5^o$

$OX = 15 \sin 24.5^o = 6.22 \ cm$

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Question 6: At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\frac{5}{12}$ on walking $192$ meters towards the tower; the tangent of the angle is found to be $\frac{3}{4}$ Find the height of the tower.

$\frac{h}{DB}$ $=$ $\frac{3}{4}$ $\Rightarrow h = DB .$ $\frac{3}{4}$

$\frac{h}{CB}$ $=$ $\frac{5}{12}$ $\Rightarrow h =$ $\frac{5}{12}$ $(192+DB)$

Therefore

$\frac{5}{12}$ $(192+DB) = DB .$ $\frac{3}{4}$

$960 + 5 DB = 9 DB \Rightarrow DB = 240 \ m$

Hence $h = 240 \times$ $\frac{3}{4}$ $= 180 \ m$

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Question 7: A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height $h$ meter. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $\alpha$ and that of the top of flagstaff is $\beta$. Prove that the height of the tower is $\frac{h \tan \alpha}{\tan \beta - \tan \alpha}$.

$\frac{h+DB}{CB}$ $= \tan \beta$

$\frac{DB}{CB}$ $= \tan \alpha$

$CB =$ $\frac{DB}{\tan \alpha}$

$\frac{h + DB}{\frac{DB}{\tan \alpha}}$ $= \tan \beta$

$h + DB = DB \Big($ $\frac{ \tan \beta}{\tan \alpha}$ $\Big)$

$\Rightarrow h = \Big($ $\frac{\tan \beta - \tan \alpha}{\tan \alpha}$ $\Big) DB$

$\therefore DB = \Big($ $\frac{\tan \alpha}{\tan \beta - \tan \alpha}$ $\Big) h$

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Question 8: With reference to the given figure of a man stands on the ground at point $A$, which is on the same horizontal plane as $B$, the foot of the vertical pole $BC$. The height of the pole is $10 \ m$. The man’s eye is $2 \ m$ above the ground. He observes the angle of elevation of $C$, the top of the pole, as $x^o$ where $\tan x = \frac{2}{5}$ Calculate: (i) the distance $AB$ in meters; (ii) angle of elevation of the top of the pole when he is standing $15$ meters from the pole. Give your answer to the nearest degree.    [1999]

$\tan x =$ $\frac{2}{5}$

$\frac{CE}{DE}$ $= \frac{2}{5}$

$DE =$ $\frac{5}{2}$ $\times ( 10 - 2) = 20$ m

i) $AB = DE = 20$ m

ii) if the distance from pole $= 15$ m

$\therefore \tan y =$ $\frac{8}{15}$ $= 0.5333 \Rightarrow y = 28.072^o$

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Question 9: The angles of elevation of the top of a tower from two points on the ground at distances $a$ and $b$ meters from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt{ab}$ meter.

$\tan \alpha =$ $\frac{h}{a}$

$\tan \beta =$ $\frac{h}{b}$

$\alpha + \beta = 90^o$

$\tan (\alpha + \beta) = \tan (90^o)$

$\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha . \tan \beta}$ $= \tan (90^o)$

$\Rightarrow 1 - \tan \alpha . \tan \beta = 0$

$\frac{h}{a} . \frac{h}{b}$ $= 1$

$\Rightarrow h^2 = ab$

$\Rightarrow h = \sqrt{ab}$

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Question 10: From a window $A$, $10 \ m$ above the ground the angle of elevation of the top $C$ of a tower is $x^o$, where $\tan x^o = \frac{5}{2}$ and the angle of depression of the foot $D$ of the tower is $y^o$ where $\tan y^o = \frac{1}{4}$. See the given figure. Calculate the height $CD$ of the tower in meters.    [2000]

$\tan x =$ $\frac{5}{2}$ $\Rightarrow$ $\frac{EC}{AE}$ $=$ $\frac{5}{2}$

$\tan y =$ $\frac{1}{4}$ $\Rightarrow$ $\frac{ED}{AE}$ $=$ $\frac{1}{4}$ $\Rightarrow AE = 4 \times 10 = 10\ m$

$\therefore EC = 40 \times$ $\frac{5}{2}$ $= 100 \ m$

$\therefore CD = ED + EC = 10 + 100 = 110 \ m$

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Question 11: A vertical tower is $20 \ m$ high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is $0.53$. How far is he standing from the foot of the tower?

$\cos \theta = 0.53$

$\Rightarrow$ $\frac{AB}{CB}$

$CB = \sqrt{20^2 + x^2}$

$\therefore x = 0.53 \sqrt{20^2 + x^2}$

$3.56 x^2 = 20^2 + x^2$

$\Rightarrow x =$ $\sqrt{ \frac{400}{2.56} }$

$\Rightarrow x = 12.5 \ m$

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Question 12: A man standing on the bank of a river observes that the angle of elevation of the top of the tower is $60^o$. When he moves $50 \ m$ away from the bank. he finds the angle of elevation to be $30^o$. Calculate: (i) the width of the river and (ii) the height of the tree.    [2003]

Let the height of the tree $= h$

Let the width of the river $= w$

$\therefore$ $\frac{h}{w}$ $= \tan 60^o = \sqrt{3}$

$\frac{h}{w+50}$ $= \tan 30^o =$ $\frac{1}{\sqrt{3}}$

$\therefore h = \sqrt{3} w$

$\Rightarrow \sqrt{3} h = w + 50$

$\Rightarrow \sqrt{3} (\sqrt{3} w) = w + 50$

$\Rightarrow 3w = w + 50$

$\Rightarrow 2w = 50$

$\Rightarrow w = 25 \ m$

$\therefore h = 25\sqrt{3} = 43.3 \ m$

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Question 13: A $20 \ m$ high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is $60^o$ and angle of elevation of the top of the pole as seen from the foot of the tower is $30^o$. Find: (i) the height of the tower (ii) the horizontal distance between the pole and the tower.

$\frac{20}{CB} = \tan 30^o =$ $\frac{1}{\sqrt{3}}$

$CB = 20\sqrt{3} \ m$

$\therefore$ $\frac{AB}{CB} = \tan 60^o = \sqrt{3}$

$\Rightarrow AB = 20 \sqrt{3} \times \sqrt{3} = 60 \ m$

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Question 14: A vertical pole and vertical tower are on the same ground level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is $60^o$ and the angle of depression of the bottom of the tower is $30^o$ Find: (i) the height of the tower, if the height of the pole is $20 \ m$;(ii)the height of the pole; if the height of the tower is $75 \ m$.

i) $\frac{DE}{AE}$ $= \tan 60 \Rightarrow DE = AE (\sqrt{3})$

$\frac{EC}{AE}$ $= \tan 30 \Rightarrow 20 \sqrt{3}$

$\therefore DE = 60$ m

Height of tower $= 60 + 20 = 80$ m

ii) $\frac{75-h}{AE}$ $= \tan 60 \Rightarrow 75 - h = \sqrt{3} AE$

$\frac{h}{AE}$ $= \tan 30 \Rightarrow AE = \sqrt{3} h$

$\therefore 75 - h = \sqrt{3} (\sqrt{3} h)$

$75 = 4 - h$

$h = 18.5$ m

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Question 15: From a point, $36 \ m$ above the surface of a lake, the angle of elevation of a bird is observed to be $30^o$ and angle of depression of its image in the water of the lake is observed to be $60^o$. Find the actual height of the bird above the surface of the lake.

$\frac{h}{BC}$ $= \tan 30 \Rightarrow BC = \sqrt{3} h$

$\frac{EF + 36}{BC}$ $= \tan 60$

$\Rightarrow EF + 36 = \sqrt{3} h ( \sqrt{3}) = 3h$

$\therefore EF = 3h - 36$

Now, $AE = EF$

$\Rightarrow h + 36 = 3h - 36$

$\Rightarrow 72 = 2h$

$\Rightarrow h = 36$ m

Therefore height of the bird $= 72$ m

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Question 16: A man observes the angle of elevation of the top of a building to be $30^o$. He walks towards it in a horizontal line through its base. On covering $60 \ m$, the angle of elevation changes to $60^o$. Find the height of the building correct to the nearest meter.

$\frac{h}{CB}$ $= \tan 60 \Rightarrow CB =$ $\frac{h}{\sqrt{3}}$

$\frac{h}{DB}$ $= \tan 30 \Rightarrow DB = \sqrt{3} h$

$\therefore \sqrt{3} h -$ $\frac{h}{\sqrt{3}}$ $= 60$

$\Rightarrow h \Big($ $\frac{3-1}{\sqrt{3}}$ $\Big) = 60$

$h = 30 \sqrt{3} = 51.96$ m $\approx 52$ m

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Question 17: As observed from the top of a, $80 \ m$ tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base, are $30^o$ and $40^o$ respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.    [2012]

$\frac{80}{BC}$ $= \tan 40 \Rightarrow$ $\frac{80}{0.839}$ $= 95.34$

$\frac{80}{BC+CD}$ $= \tan 30$

$\Rightarrow 80\sqrt{3} = 95.34 + CD$

$CD = 43.22$ m $\approx 43$ m

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Question 18: In the given figure, from the top of a building $AB = 60 \ m$ high, the angles of depression of the top and bottom of a vertical lamp post $CD$ are observed to be $30^o$ and $60^o$ respectively. Find: (i) the horizontal distance between $AB$ and $CD$. (ii) the height of the lamp Post.

i) $\frac{60}{BC}$ $= \tan 60 \Rightarrow BC =$ $\frac{60}{\sqrt{3}}$ $= 20 \sqrt{3} = 34.34$ m

ii) $\frac{60-h}{20\sqrt{3}}$ $= \tan 30 \Rightarrow$ $\frac{60-h}{20\sqrt{3}}$ $=$ $\frac{1}{\sqrt{3}}$ $\Rightarrow h = 40$ m

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Question 19: An airplane, at an altitude of $250 \ m$ observes the angles of depression of two boats on the opposite banks of a river to be $45^o$ and $60^o$ respectively. Find the width of the river. Write the answer correct to the nearest whole number.    [2014]

$\frac{250}{BD}$ $= \tan 45 \Rightarrow BD = 250$ m
$\frac{250}{CD}$ $= \tan 60 \Rightarrow CD =$ $\frac{250}{\sqrt{3}}$
$= BD + CD = 250 +$ $\frac{250}{\sqrt{3}}$ $= 250 + 144.34 = 394.34$ m
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