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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADB  

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

Substituting from above

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

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

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

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADC  

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

Substituting from above

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

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

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

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADB  

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

Substituting from above

\displaystyle h = (40+ h) \tan 30^o  

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

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

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Question 4: From the top of a light house \displaystyle 100\text{ m } high, the angles of depression of two ships are observed as \displaystyle 48^o and \displaystyle 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]

Answer:

(i) \displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADB  

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

Therefore, the dis\tance between the ships is =

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

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

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

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

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

\displaystyle = 90.040 + 136.63 = 227.68 \approx 228\text{ m }  

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle DEC  

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

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

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

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

Hence \displaystyle h = 37.5 \times \tan 60^o = 64.95\text{ m }  

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle AED  

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

\displaystyle \text{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 \displaystyle 50^o . At a Point, \displaystyle 30\text{ m } vertically above the first point of observation, the elevation is found to be \displaystyle 45^o Find: (i) the height of the tower, (ii) its horizontal distance from the points of Observation

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADE  

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

\displaystyle \text{Therefore } h = (h - 30 ) \tan 60^o  

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

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

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

\displaystyle \\

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle AED  

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

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

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

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Question 9: A man on a cliff observes a boat. at an angle of depression \displaystyle 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 \displaystyle 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 \displaystyle 500\text{ m } .

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ABD  

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

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

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

\displaystyle \Rightarrow x = 577.35\text{ m }  

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

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

Speed of the boat \displaystyle = \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 \displaystyle 150\text{ m } high, takes \displaystyle 2 minutes to change the angle of elevation of the top of the lighthouse \displaystyle \text{From } 60^o to \displaystyle 45^o . Find the speed of the boat.

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ABD  

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

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

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

\displaystyle \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 \displaystyle 60^o . When he moves \displaystyle 40\text{ m } away from the bank, he finds the angle of elevation to be \displaystyle 30^o . Find: (i) the height of the free, correct to \displaystyle 2 decimals places, (ii) the width of the river.

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ABD  

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

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

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

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

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

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \Rightarrow 160 - h = 75  

\displaystyle \Rightarrow h = 160 - 75 = 85\text{ m }  

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

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ABD  

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

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

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

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

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

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

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

Answer:

\displaystyle \text{From } \triangle EDC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ADB  

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

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

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

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

\displaystyle \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 \displaystyle 30^o and \displaystyle 45^o respectively. Find the distances of the two stones from the foot of the hill. [2007]

Answer:

\displaystyle \text{From } \triangle ABC  

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

\displaystyle \text{Similarly, } \text{From } \triangle ABD  

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

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

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

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

\displaystyle \text{Therefore } BC = 1.366 \text{ km and } DC = 2.366 \text{ km }