Question

The angle of elevation of the top of a tower from two points at a distance of 100m and 150 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 50√6m

Answer 1

• Let the height of the tower (CD) = h m
• AC be a horizontal line on a ground .
• A and B be the two points on a line at a distance of 150 m and 100 m from the base of the tower.
• $Let \: \angle {CBD } = \theta; \:then \\ \angle {CAD} = 90 - \theta$
• ( The complementary means the sum of two angles is 90° )

$In \:right\:angled \: \triangle ACD, \\tan (90- \theta )= \frac{CD}{AC}$

$\implies cot \theta = \frac{h}{150} \:---(1)$

$In \:right\:angled \: \triangle BCD, \\tan \theta = \frac{CD}{BC}$

$\implies tan \theta = \frac{h}{100} \:---(2)$

/* From Multiplying (1) and (2) , we get */

$cot \theta \times tan \theta = \frac{h}{150} \times \frac{h}{100}$

$\implies 1 = \frac{h^{2}}{150\times 100}$

$\implies h^{2} = 6 \times 25 \times 100$

$\implies h = \sqrt{6 \times 25 \times 100}$

$\implies h = \sqrt{6 \times 5^{2} \times 10^{2}}$

$\implies h =50 \sqrt{6}\:m$

Therefore.,

$\red { Height \:of \:the \:tower }\green { =50 \sqrt{6}\:m}$

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