Question

on a straight line passing through the foot of the tower two points. C and D are a distance of 4 m and 16 m from the foot respectively if the angle of elevation of c and d from the top of the tower are complementary then find the height of the tower

Answer 1

Answer:

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Given :-

• Let the angle of elevation /_ACB & /_ADB be θ and 90° - θ.
• Let AC = 4 m and AD = 16 m.
• Let 'h' be the height of the tower AB.

To Find :-

• Height of tower 'h'.

Formula used :-

◇ tan (90° - θ ) = cot θ

◇ cot θ = 1/ tanθ

Solution:-

Now, In right triangle ABC

$\large\bold\red{tanθ \: = \frac{AB}{AC} }$

$\large\bold\red{tanθ = \frac{h}{4} }......(1)$

Now, In right triangle ADB

$\large\bold\red{tan(90 - θ ) = \frac{AB}{AD} }$

$\large\bold\red{cotθ = \frac{h}{16} }.....(2)$

Multiply (1) and (2), we get

$\large\bold\red{tanθ \times cot θ = \frac{h}{4} \times \frac{h}{16} }$

$\large\bold\red{tanθ \times \frac{1}{tanθ } = \frac{h}{4} \times \frac{h}{16} }$

$\large\bold\red{1 = \frac{ {h}^{2} }{64} }$

$\large\bold\red{ {h}^{2} = 64}$

$\large\bold\red{⟹ \: h \: = 8 \: m}$