Question

20 metre high flag pole is fixed on a80m high pillar 50 metre high away from it on a point on the base of pillar the flag pole makes an angle alpha then the value of tan alpha

Answer 1

Given:

A 20-metre high flag pole is fixed on an 80-metre high pillar,

50 m away from it, on a point on the base of the pillar, the flag pole makes an angle alpha

To find:

The value of tan alpha

Solution:

Referring to the figure attached below, let's assume,

AB = length of the flag pole = 20 m

BC = length of the pillar = 80 m

CD = distance between the base of the pillar and the point D

∠ADB = α = angle made by the flag pole on the point D

∠BDC = β = angle made by the pillar at point D

Considering Δ BCD, we have

$tan \:\beta = \frac{perpendicular \:height}{base}$

$\implies tan \:\beta = \frac{BC}{CD}$

$\implies tan \:\beta = \frac{80}{50}$

$\implies \bold{tan \:\beta = \frac{8}{5}}$ ..... (i)

Now, considering Δ ACD, we have

$tan (\alpha +\:\beta) = \frac{perpendicular \:height}{base}$

$\implies tan (\alpha + \:\beta) = \frac{AC}{CD}$

$\implies tan (\alpha + \:\beta) = \frac{AB + BC}{CD}$

$\implies tan (\alpha + \:\beta) = \frac{20 + 80}{50}$

$\implies tan (\alpha + \:\beta) = \frac{100}{50}$

$\implies tan (\alpha + \:\beta) = 2$

$\implies \frac{tan\:\alpha \:+\: tan\:\beta}{1 \:-\: tan\:\alpha tan\:\beta} = 2$

substituting from (i), we get

$\implies \frac{tan\:\alpha \:+\: \frac{8}{5} }{1 \:-\: tan\:\alpha \times \frac{8}{5} } = 2$

$\implies \frac{5tan\:\alpha \:+\: 8 }{5 \:-\: 8tan\:\alpha } = 2$

$\implies 5tan\:\alpha \:+\: 8 = 2(5 \:-\: 8tan\:\alpha )$

$\implies 5tan\:\alpha \:+\: 8 = 10 \:-\: 16tan\:\alpha$

$\implies 16tan\:\alpha + 5tan\:\alpha = 10 \:\:-\: 8$

$\implies 21tan\:\alpha = 2$

$\implies \bold{tan\:\alpha = \frac{2}{21} }$

Thus, $\boxed{\bold{The\:value\:of\:tan\:\alpha\:is \:\underline{\frac{2}{21} }}}.$

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