[CBSE Topper 2014, 2018]
From a point on the ground, the angles of elevation of the bottom and the top of a transmission 60° respectively. Find the height of the tower fixed at the top of a 20 m high building are 45° and tower. (Use √7 = 1-73)
From a point on the ground, the angles of elevation of the bottom and the top of a transmission 60° respectively. Find the height of the tower fixed at the top of a 20 m high building are 45° and tower. (Use √7 = 1-73)
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Let the height of the building is BC, the height of the transmission tower which is fixed at the top of the building be AB.
D is the point on the ground from where the angles of elevation of the bottom B and the top A of the transmission tower AB are 45° and 60° respectively.
The distance of the point of observation D from the base of the building C is CD.
Combined height of the building and tower = AC = AB + BC
Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θ.
In ΔBCD,
tan 45° = BC/CD
1 = 20/CD
CD = 20
In ΔACD,
tan 60° = AC/CD
√3 = AC/20
AC = 20√3
Height of the tower, AB = AC - BC
AB = 20√3 - 20 m
= 20 (√3 - 1) m
Answer:
Let the height of the building is BC, the height of the transmission tower which is fixed at the top of the building be AB.
D is the point on the ground from where the angles of elevation of the bottom B and the top A of the transmission tower AB are 45° and 60° respectively.
The distance of the point of observation D from the base of the building C is CD.
Combined height of the building and tower = AC = AB + BC
Trigonometric ratio involving sides AC, BC, CD, and ∠D (45° and 60°) is tan θ.
In ΔBCD,
tan 45° = BC/CD
1 = 20/CD
CD = 20
In ΔACD,
tan 60° = AC/CD
√3 = AC/20
AC = 20√3
Height of the tower, AB = AC - BC
AB = 20√3 - 20 m
= 20 (√3 - 1) m
I hope it will help you