The angle of elevation of the top of a tower from two points P and Q at a distance of ‘a’ and ‘b’ respectively, from the base and in the same straight line with it are complementary. Prove that height
Answer: A
Let L SPR = θ then L SQR = 90º – θ and RS = h,
In Δ SPR, tan θ = h/a Or, h = a tan θ --------------------------------- (i)
In ΔSQR, tan (90º – θ) = h/b Or, h = b tan (90º – θ) = h cot θ ----- (ii)
Multiplying (i) and (ii) we get, h2 = a tan θ b cot θ = a b
Or, h = √(a b). [Proved.]
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From the foot and the top of a building of height 230 m, a person observes the top of a tower with angles of elevation of b and a respectively. What is the distance between the top of these buildings if tan a = 5/12 and tan b = 4/5
Answer: B
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The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
Answer: D
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A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how
Answer: C
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A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?
Answer: C
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From a window A, 10 m above ground the angle of elevation of the top C to a tower is xº , where tan xº = 5/2 and the angle of depression of the foot D of the tower is yº , where tan yº = 1/4 . Calculate
Answer: C
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When height of a tree is equal to the length of its shadow, what is the angle of elevation of the sun?
Answer: D
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A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60º. When he moves 50 m away from the bank, he finds the angle of elevation to be 30º.
Answer: D
Fig.
Let AD be the tree of height h,
In ΔADC,
tan60º = h/CD
Or, √3 = h/CD
Or, CD = h/√3
In ΔADB,
tan30º = h/BD
Or, 1/√3 = h/BD
Or, BD = h√3
BD – CD = 50
h√3 – h/√3 = 50
(3h – h)/√3 = 50
2h = 50√3
Height of the tree,
h = 50√3/2
= 25√3 25×1.732
= 43.3 m.
Width of the river,
CD = h/√3
= 25√3/√3
= 25 m.
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The thread of a kite is 120 m long and it is making 30° angular elevation with the ground .What is the height of the kite?
Answer: A
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From the top of a hill, the angle of depression of two consecutive kilometer stones, due east are found to be 30º and 45º respectively. Find the distance of the two stones from the foot of the hill.
Answer: A
Fig.
Let AB be the hill whose foot is B and D and C are two kilometer stones.
Therefore DC = 1 km = 1000 m,
Let AB = h and BC = x.
In right angled Δ ABC,
tan 45º = AB/BC
=> 1 = h/x
=> x = h ------------- (1)
In right angled Δ ABC,
tan 30º = AB/BD
=> 1/√3 = h/(x + 1000)
=> x + 1000 = h√3 ----------------(2)
using (1) and (2) we get
x + 1000 = x√3
=> x(√3 – 1) = 1000
Or, x = [1000/(√3 – 1)] × (√3 + 1)/(√3 + 1)
Or, x = 1000 (√3 + 1)/2
= 500(√3 + 1)
= 500 × 2.732
= 1366 m or 1.366 km.
Therefore first km stone is 1.366 km and second km stone is 2.366 km from the foot of the hill.
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