Question

if the sun's ray inclination increases from 45degree to 60degree,the length of shadow of a tower decreases by 50m. height of the tower (in m) is.

Answer 1

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Answer 2

Answer:

118.2 m

Step-by-step explanation:

Refer the attached figure .

Height of tower Ab = h

∠ACB = 60°

∠ADB = 45°

The length of shadow of a tower decreases by 50m i.e.CD = 50 m

Let BC = x

So, BD = x+50

In ΔABC

Using trigonometric ratios

$tan\theta = \frac{Perpendicular}{Base}$

$tan 60 ^{\circ} = \frac{AB}{BC}$

$\sqrt{3} = \frac{h}{x}$

$\sqrt{3}x = h$   -1

In ΔABD

Using trigonometric ratios

$tan\theta = \frac{Perpendicular}{Base}$

$tan 45 ^{\circ} = \frac{AB}{BD}$

$1= \frac{h}{x+50}$

$x+50 = h$   --2

Equating 1 and 2

$x+50 = \sqrt{3}x$

$50 = \sqrt{3}x-x$

$50 =x(\sqrt{3}-1)$

$50 =x(0.7320)$

$\frac{50}{0.7320} =x$

$68.3 =x$

Substitute the values of x in 1

$\sqrt{3}(68.3) = h$

118.2 = h

Hence the height of the tower is 118.2 m