Height and Distance Questions

FACTS  AND  FORMULAE  FOR  HEIGHT  AND  DISTANCE  PROBLEMS

 

 

1.In a right angled OAB, where BOA=θ

 

(i). sinθ =PerpendicularHypotenuse=ABOB

 

(ii). cosθ=BaseHypotenuse=OAOB

 

(iii). tanθ=PerpendicularBase=ABOA

 

(iv). cosecθ =1sinθ=OBAB

 

(v). secθ=1cosθ=OBOA

 

(vi). cotθ=1tanθ=OAAB

 

2. Trigonometrical Identities :

(i)sin2θ+cos2θ=1

(ii). 1+tan2θ=sec2θ

(iii). 1+cot2θ=cosec2θ

 

3. Values of T-ratios :

θ0o30o45o60o90osin θ01212321cos θ13212120tan θ01313Not defined

 

4. Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

Therefore, Angle of elevation of P from O is =AOP

 

5. Angle of Depression :

Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Q:

From a point P, the angle of elevation of a tower is such that its tangent is 3/4. On walking 560 metres towards the tower the tangent of the angle of elevation of the tower becomes 4/3. What is the height (in metres) of the tower?

A) 720 B) 960
C) 840 D) 1030
 
Answer & Explanation Answer: B) 960

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Q:

The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60° . The height of the tower is

A) 5(√3 - 1) m B) 5(√3 + 1) m
C) 10 (√3 - 1) m D) 10 (√3 + 1) m
 
Answer & Explanation Answer: B) 5(√3 + 1) m

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Q:

From the top of a 20 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of it's foot is at 45°, then the height of the tower is (√3=1.732)

A) 45.46 m B) 45.64 m
C) 54.64 m D) 54.46 m
 
Answer & Explanation Answer: C) 54.64 m

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Q:

A tower is 50 meters high.Its shadow is x metres shorter when the sun's altitude is 45° than when it is 30° . The value of x in metres is

A) 50√3 B) 50 (√3 -­ 1)
C) 50 (√3 + 1) D) 50
 
Answer & Explanation Answer: B) 50 (√3 -­ 1)

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Q:

From the top of a tower 60 mts high the angle of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. If the pole and tower stand on the same plane, the height of the pole in meters is

A) 60(√3­1) B) 20(√3­1)
C) 20(3­√3) D) 20(√3+1)
 
Answer & Explanation Answer: C) 20(3­√3)

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Q:

Two trees are standing along the opposite sides of a road. Distance between the two trees is 400 metres. There is a point on the road between the trees. The angle of depressions of the point from the top of the trees are 45 deg and 60 deg. If the height of the tree which makes 45 deg angle is 200 metres, then what will be the height (in metres) of the other tree?

A) 200 B) 200√3
C) 100√3 D) 250
 
Answer & Explanation Answer: B) 200√3

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Q:

The angle of elevation of the top of a hill at the foot of the tower is 60 deg and the angle of elevation of the top of the tower from the foot of the hill is 30 deg. If the tower is 50 m high, what is the height of the hill?

A) 100 m B) 120 m
C) 180 m D) 150 m
 
Answer & Explanation Answer: D) 150 m

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Q:

A boat is moving away from an observation tower. It makes an angle of depression of 60° with an observer's eye when at a distance of 50m from the tower. After 8 sec., the angle of depression becomes 30°. By assuming that it is running in still water, the approximate speed of the boat is

A) 33 km/hr B) 42 km/hr
C) 45 km/hr D) 50 km/hr
 
Answer & Explanation Answer: C) 45 km/hr

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