Question # 1

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60° and when he retires 40 meters away from the tree the angle of elevation becomes 30°. The breadth of the river is

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40 m

30 m

20 m

60 m

A tower subtends an angleαat a point on the same level as the root of the tower and at a second point, b meters above the first, the angle of depression of the foot of the tower isβ. The height of the tower is

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b cotαtanβ

b tanαtanβ

b tanαcotβ

None of these

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E-radius corresponding to < B is

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The angle of elevation of the tops of two towers at the middle point of the line joining the foots of the tower are 60^οand 30^οrespectively. The the ratio of the heghts of the tower is

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2 : 1

3 : 1

1 : 2

1 : 3

The angle of depression of a point situated at a distance of 70 meters from the base of a tower is 45°. The height of the tower is

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70 m

85 m

35 m

None of these

PQ is a post of given height a, and AB is a tower at some distance;αandβare the angles of elevation of B, the top of the tower, at P and Q respectively. The height of the tower and its distance from the post are

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30°

60°

45°

None of these

If Cosθ=0, thenθ= ______

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nπ/2

(2n + 1)π/2

(2n - 1)π/2

(4n + 1)π/2

A tower subtends an angle of 30° at a point distant d from the foot of the tower and on the same level as the foot of the tower. At a second point, h vertically above the firs, the angle of depression of the foot of the tower, is 60°. The height of the tower is

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h/3

h/3d

3h

3h / d

Ifθ= 60° then

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sinθ= 1/2

tanθ= cot 30°

θ=π/4

secθ= 4

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If the flag-staff 6 meters high placed on the top of a tower. Makes the shadow 2√3 m on the ground, then the angle of elevation of the sun is

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30^ο

35^ο

45^ο

60^ο

When the angle between the ground and the sun is 30^𝜊,flag pole costss a shadow of 40 mg long. the height of the top of the flag is

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25m

23m

12m

29m

The angle of elevation of a tower from a point A due south of it is x and from a point B due east of A is y. If AB = 1, then the height h of the tower is given by

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The law of consines

The law of sines

The law of tangents

None of these

The quadratic equation 8 sec2θ - 6 secθ +1 =0 has

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Infinitely many roots

Exactly two roots

Exactly four roots

No roots

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A circle drawn inside a triangle and touching its sides is called

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In-circle

Circum circle

Escribed circle

None of these

The law of sines can be used to solve

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Right angle triangle

Isosceles triangle

oblique triangle

haxagon

An airplane flying at height of 300 meters above the ground passes vertically above another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Then the height of the lower plane from the ground is (in meters).

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For any equilateral r :R :η :r1 :r2 :r3 =

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1:2:3:4:5

1:2:3:3:3

1:2:4:4:4

2:1 :2 :2 :2

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The towers each 120 meters high are 800 meters apart. The measure of the angle of elevation from the base of one tower to the top of the other is

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12^𝜊

9^𝜊

7^𝜊

-120^𝜊

The law of sines can be used to solve oblique triangle when following information is given:

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Two angles and a side

Two sides and an angle opposite one of the given sides

Two sides and the angle between two sided

Option a and b

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AB is a vertical pole and C is its middle point. The end A is on the level ground and P is any point on the level ground other than A. the portion CB subtends and angleβat P. If AP : AB = 2 : 1 thenβ=

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Area ofΔABC=

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ab sinα

1/2 ab sinα

1/2 ac sinγ

1/2 ac sinβ

If you are looking a high point from the ground, then the angle formed is

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Angle of elevation

Angle of depression

Right angle

Horizon

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