A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30^{o}. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60^{o}. Then the time taken (in minutes) by him, from B to reach the pillar, is:
6
10
20
20
If the angles of elevation of the top of a tower from three collinear points A, B and C on line leading to the foot of the tower are 30^{o}, 45^{o} and 60^{o} respectively, then the ratio AB: BC is
ABCD is a trapezium such that AB and CD are parallel and BC⊥CD. If ∠ADB = θ, BC = p and CD = q,then AB is equal to
The expression can be written as
sin A cos A+1
sec A cosec A+1
tan A + cot A
tan A + cot A
Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB.If ∠BPC = β , then tanβ is equal to
4/9
6/7
1/4
1/4
Let A and B denote the statements
A: cos α + cosβ + cosγ = 0
B : sinα + sinβ + sinγ = 0
If cos(β – γ) + cos(γ – α) + cos(α – β) = – 3/2, then
A is true and B is false
A is false and B is true
both A and B are true
both A and B are true
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is 45°.
Then the height of the pole is