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प्रश्न
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height `10sqrt(3)` m
उत्तर
Height of the tower (AC) = `10sqrt(3)` m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ABC, tan θ = `"AC"/"AB"`
= `(10sqrt(3))/30`
= `sqrt(3)/3`
tan θ = `1/sqrt(3)`
= tan 30°
∴ Angle of inclination is 30°
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