The angle of elevation of a building from two points on the ground is 9m and 16m away from the foot of the building are complementary. Find the height of the building

Let AB be the tower and is equal to h metres D and C are the points which are 4m and 9m away from the foot of the tower.

Let ∠ACB = θ and ∠ADB = 90 – θ         (since the angles are complementary)

$\frac{AB}{BD}=\mathrm{tan}\left(90-\theta \right)\phantom{\rule{0ex}{0ex}}⇒\frac{h}{9}=cot\theta$    ......(1)

In ΔABC

$\frac{AB}{BC}=\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\frac{h}{16}=tan\theta \phantom{\rule{0ex}{0ex}}$     .....(2)

multiply equation (1) & (2)

Hence, height of the building is 12 m.

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