A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower? (2001)
Solution:
Let AB be the tower and
let a man C stands at a distance from the foot of the tower = x m
and cos θ = 0.53
The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38°30′ at a point 6 m from the foot of the tree. Calculate.
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
Solution:
Let TR be the total height of the tree
and TP is the broken part which touches the ground
at the distance of 6 m from the foot of the tree
making an angle of 38°30′ with the ground.
Let PR = x and TR = x + y
PQ = PT = y
In right ∆PQR
Height of the tree = 4.7724 + 7.6665 = 12.4389 = 12.44 m
and height of the tree at which it is broken = 4.77 m
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