Q.Two points on the ground lie along the same straight line with the base of a vertical pillar. From these two points, the angles of elevation to the top of the pillar are complementary. If the distances from the two points to the base of the pillar are 9 meters and 16 meters respectively, and both points are on the same side of the pillar, find the height of the pillar.

Let AB be the vertical pillar, and points C and D lie on the same straight line on the ground with the base B of the pillar. From points C and D, the angles of elevation to the top of the pillar A are ∠ACB and ∠ADB, which are complementary. Given: BC = 9 meters and BD = 16 meters. Let ∠ACB = \(\theta\) Then, ∠ADB = \(90^\circ - \theta\) From triangle ABC: \[ \tan \theta = \frac{AB}{BC} = \frac{AB}{9} \quad \text{(i)} \] From triangle ABD: \[ \tan(90^\circ - \theta) = \frac{AB}{BD} \Rightarrow \cot \theta = \frac{AB}{16} \Rightarrow \tan \theta = \frac{16}{AB} \quad \text{(ii)} \] Equating equations (i) and (ii): \[ \frac{AB}{9} = \frac{16}{AB} \Rightarrow AB^2 = 144 \Rightarrow AB = \sqrt{144} = 12 \] Therefore, the height of the pillar is 12 meters.
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