Question 1: In the figure, given below, it is given that $AB$ is perpendicular to $BD$ and is of length $X$ meters. $DC = 30 \ m, \angle ADB = 30^o$ and $\angle ACB = 45^o$ Without using tables, find $X$.

From $\triangle ABC$

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

Similarly, from $\triangle ADB$

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

Substituting from above

$x = 30 \ \ 30^o + x \ tan \ 30^o$

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

$\Rightarrow x =$ $\frac{30 \ tan \ 30^o}{1- tan \ 30^o}$ $= 40.98 \ m$

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Question 2: Find the height of a tree when it is found that on walking away from it $20 \ m$? in a horizontal line through its base, the elevation of its top changes from $60^o$ to $30^o$.

From $\triangle ABC$

$\frac{x}{BC}$ $= tan \ 60^o \Rightarrow BC =$ $\frac{x}{tan \ 60^o}$

Similarly, from $\triangle ADC$

$\frac{x}{20+BC}$ $= tan \ 30^o$

Substituting from above

$x = (20 +$ $\frac{x}{tan \ 60^o})$ $tan \ 30^o$

$x(1-$ $\frac{tan \ 30^o}{tan \ 60^o}$ $) = 30 \ tan \ 30^o$

$\Rightarrow x =$ $\frac{11.547}{0.6667}$ $= 17.32 \ m = 40.98 \ m$

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Question 3: Find the height of a building, when it is found that on walking towards it $40 \ m$ in a horizontal line through its base the angular elevation of its top changes from $30^o$ to $45^o$.

From $\triangle ABC$

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

Similarly, from $\triangle ADB$

$\frac{h}{40+BC}$ $= tan \ 30^o$

Substituting from above

$h = (40+ h) \ tan \ 30^o$

$h\ (1- tan \ 30^o) = 40 \ tan \ 30$

$\Rightarrow h =$ $\frac{40 \ tan \ 30^o}{1- tan \ 30^o}$ $= 54.64 \ m$

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Question 4: 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 5: Two pillars of equal heights stand on either side of a roadway, which is $150 \ m$ wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are $60^o$ and $30^o$; find the height position of the pillar and the position of the point.

From $\triangle ABC$

$\frac{h}{x}$ $= tan \ 60^o$

Similarly, from $\triangle DEC$

$\frac{h}{150-x}$ $= tan \ 30^o$

$\therefore x \ tan \ 60^o = (150 - x) \ tan \ 30^o$

$x \ (tan \ 60^o + tan \ 30^o) = 150 \ tan \ 30^o$

$x =$ $\frac{150 \ tan \ 30^o}{tan \ 60^o+tan \ 30^o}$ $= 37.5 \ m$

Hence $h = 37.5 \times tan \ 60^o = 64.95 \ m$

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Question 6: From the figure, given below, calculate the length of $CD$.

From $\triangle ABC$

$\frac{AB}{15}$ $= tan \ 47^o$

Similarly, from $\triangle AED$

$\frac{AB}{15}$ $= tan \ 22^o$

Therefore $CD = AB - AE = 15 \ tan \ 47^o - 15 \ tan \ 22^o = 16.085-6.060 = 10.025$

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Question 7: The angle of elevation of the top of a tower is observed to be $50^o$. At a Point, $30 \ m$ vertically above the first point of observation, the elevation is found to be $45^o$ Find: (i) the height of the tower, (ii) its horizontal distance from the points of Observation

From $\triangle ABC$

$\frac{h}{BC}$ $= tan \ 60^o$

Similarly, from $\triangle ADE$

$\frac{h - 30}{BC}$ $= tan \ 45^o \Rightarrow BC = (h - 30)$

Therefore $h = (h - 30 ) \ tan \ 60^o$

Given $(h - 30 )\ tan \ 60^o = -30 \ tan \ 60^o$

$\Rightarrow h =$ $\frac{-30 \ tan \ 60^o}{1 - tan \ 60^o}$ $= 70.98 \ m$

Therefore $BC =$ $\frac{h}{ tan \ 60^o} = \frac{70.98}{ tan \ 60^o}$ $= 40.98 \ m$

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Question 8: From the top of a cliff, $60 \ m$ high, the angles of depression of the top and bottom of a tower are observed to be $30^o$ and $60^o$. Find the height of the tower.

From $\triangle ABC$

$\frac{60}{BC}$ $= tan \ 60^o$

Similarly, from $\triangle AED$

$\frac{60- h}{BC}$ $= tan \ 30^o$

Therefore $60 - h =$ $\frac{60}{tan \ 60^o}$ $. tan \ 30^o$

$\Rightarrow h = 60 - 60 ($ $\frac{tan \ 30^o}{tan \ 60^o})$ $= 40 \ m$

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Question 9: A man on a cliff observes a boat. at an angle of depression $30^o$, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be $60^o$. Assuming that the boat sails at a uniform speed, determine: (i) how much more time it will take to reach the shore. (ii) the speed of the boat in meter per second if the height of the cliff is $500 \ m$.

From $\triangle ABC$

$\frac{500}{BC}$ $= tan \ 60^o \Rightarrow BC =$ $\frac{500}{tan \ 60^o}$

Similarly, from $\triangle ABD$

$\frac{500}{x+BC}$ $= tan \ 30^o$

$\therefore 500 = (x +$ $\frac{500}{tan \ 60^o}$ $) tan \ 30^o$

$\Rightarrow 500 (1 -$ $\frac{tan \ 30^o}{tan \ 60^o}$ $) = x tan \ 30^o$

$\Rightarrow x = 577.35 \ m$

Also $BC =$ $\frac{500}{tan \ 60^o}$ $= 288.675 \ m$

Therefore the time that the boat will take to reach the shore $=$ $\frac{288.675}{577.35}$ $\times 3 = 1.5 \ minutes.$

Speed of the boat $=$ $\frac{577.35}{3 \times 60}$ $= 3.21 \frac{m}{s}$

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Question 10: A man in a boat rowing away from a lighthouse $150 \ m$ high, takes $2$ minutes to change the angle of elevation of the top of the lighthouse from $60^o$ to $45^o$. Find the speed of the boat.

From $\triangle ABC$

$\frac{150}{BC}$ $= tan \ 60^o$

Similarly, from $\triangle ABD$

$\frac{150}{x+BC}$ $= tan \ 45^o$

$\therefore 150 = x +$ $\frac{150}{tan \ 60^o}$

$\Rightarrow x = 150 \ (1 -$ $\frac{1}{tan \ 60^o}$ $) = 63.34 \ m$

$\therefore Speed of the boat =$ $\frac{63.34}{2 \times 60}$ $= 0.53 \frac{m}{s}$

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Question 11: A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is $60^o$. When he moves $40 \ m$ away from the bank, he finds the angle of elevation to be $30^o$. Find: (i) the height of the free, correct to $2$ decimals places, (ii) the width of the river.

From $\triangle ABC$

$\frac{h}{BC}$ $= tan \ 60^o$

Similarly, from $\triangle ABD$

$\frac{h}{40+BC}$ $= tan \ 30^o$

$\therefore h = tan \ 30^o ($ $\frac{h}{tan \ 30^o}$ $+ 40)$

$\Rightarrow h (1 -$ $\frac{tan \ 30^o}{tan \ 60^o}$ $) = tan \ 30^o \times 40$

$h =$ $\frac{tan \ 30^o \times 40}{(1 - \frac{tan \ 30^o}{tan \ 60^o})}$

Therefore $BC =$ $\frac{34.64}{tan \ 60^o}$ $= 20 \ m$

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Question 12: The horizontal distance between two towers is $5 \ m$ and the angular depression of the top of the first tower as seen from the top of the second. which is $160 \ m$ high, is $45^o$. Find the height of the first tower.

From $\triangle ABC$

$\frac{160-h}{75}$ $= tan \ 45^o$

$\Rightarrow 160 - h = 75$

$\Rightarrow h = 160 - 75 = 85 \ m$

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Question 13: The length of the shadow of a tower standing on level plane is found to be $2y$ meters longer when the sun’s altitude is $30^o$ then when it is $45^o$. Prove that the height of the tower is $y(\sqrt{3}+1)$ meters.

From $\triangle ABC$

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

Similarly, from $\triangle ABD$

$\frac{h}{h + 2y}$ $= tan \ 30^o$

Substituting $h = 2y$ $\frac{tan \ 30^o}{1- tan \ 30^o}$

$\Rightarrow h = 2y$ $\frac{1}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3} -1})$

$\Rightarrow h = 2y$ $(\frac{1}{\sqrt{3} -1})$

Multiplying both numerator and denominator by $(\sqrt{3}+1)$ we get

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

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Question 14: After $10$ seconds, its elevation is observed to be $30^o$; find the uniform speed of the Aeroplane in km per hour.

From $\triangle EDC$

$\frac{1}{BC}$ $= tan \ 60^o \Rightarrow DC = \frac{1}{\sqrt{3}}$

Similarly, from $\triangle ADB$

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

$\Rightarrow 1 = tan \ 30^o ($ $\frac{1}{\sqrt{3}}$ $+BC)$

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

$\Rightarrow BC =$ $\frac{2}{\sqrt{3}}$

$\therefore Speed =$ $\frac{\frac{2}{\sqrt{3}}}{\frac{10}{3600}}$ $= 415.69$ $\frac{km}{hr}$

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Question 15: 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 $BC = 1.366 \ km \ and \ DC = 2.366 \ km$

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