249Question 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 .

Answer:

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 .

31Answer:

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 .

32Answer:

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]

33Answer:

(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.

34Answer:

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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248Question 6: From the figure, given below, calculate the length of CD .

Answer:

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

36Answer:

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.

37Answer:

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 .

38Answer:

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.

39Answer:

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.

310Answer:

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.

315.jpgAnswer:

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.

312Answer:

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.

313Answer:

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]

314Answer:

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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