Question 1: 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 2: 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 3: 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 4: 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 5: 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

$\frac{250}{CD}$ $= \tan 60 \Rightarrow CD =$ $\frac{250}{\sqrt{3}}$

Therefore the width of the river

$= BD + CD = 250 +$ $\frac{250}{\sqrt{3}}$ $= 250 + 144.34 = 394.34$ m

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Question 6: From the top of a light house $100 \ m$ high, the angles of depression of two ships are observed as $48^o$ and $36^o$ respectively. Find the distance between the two ships (in the nearest meter) if:

(i) the ships are on the same side of the light house,

(ii) the ships are on the opposite sides of the light house. [2010]

(i) From $\triangle ABC$

$\frac{100}{BC}$ $= tan \ 48^o \Rightarrow BC =$ $\frac{100}{tan \ 48^o}$

Similarly, from $\triangle ADB$

$\frac{100}{DB}$ $= tan \ 36^o \Rightarrow DB =$ $\frac{100}{tan \ 36^o}$

Therefore, the distance between the ships is =

$DB - BC =$ $\frac{100}{tan \ 36^o} - \frac{100}{tan \ 48^o}$

$= 100 ($ $\frac{tan \ 48^o - tan \ 36^o}{ tan \ 48^o . tan \ 36^o})$

$=$ $\frac{100 \times 0.3841}{0.8069}$ $= 47.60 \approx 48 \ m$

(ii) If the ships were on the opposite sides, then the distance between the ships is =

$DB + BC =$ $\frac{100}{tan \ 36^o} + \frac{100}{tan \ 48^o}$

$= 90.040 + 136.63 = 227.68 \approx 228 \ m$

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Question 7: From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be $30^o$ and $45^o$ respectively. Find the distances of the two stones from the foot of the hill. [2007]

From $\triangle ABC$

$\frac{h}{BC}$ $= tan \ 45^o \Rightarrow h = BC$

Similarly, from $\triangle ABD$

$\frac{h}{1 + BC}$ $= tan \ 30^o$

$BC = (BC + 1) \ tan \ 30^o$

$BC \ ( 1 - tan \ 30^o) = tan \ 30^o$

$BC =$ $\frac{tan \ 30^o}{1 - tan \ 30^o}$ $=$ $\frac{1}{\sqrt{3} - 1}$ $= 1.366 \ km$

Therefore $BC = 1.366 \ km \ and \ DC = 2.366 \ km$