Question

a hedgehog wishes to cross a road without to cross a road without being run over. he observes the angle of elevation of a lamp post on the other side of the road to be angle 45 from the edge of the road and angle 30 from a point 10 m back from the road

Answer 1

Given:

The hedgehog observes the angle of elevation of a lamp post on the other side of the road to be angle 45° from the edge of the road and angle 30° from a point 10 m back from the road

To find:

How wide is the road?

Solution:

To solve the given problem we will use the following formula:

$\boxed{\bold{tan\theta = \frac{Opposite \:side}{Adjacent\:side} }}$

Referring to the figure attached below we have,

AB = height of the lamp post

BC = width of the road

CD = distance covered by the hedgehog back from the road = 10 m

∠ACB = 45° = angle of elevation of a lamp post on the other side of the road

∠ADB = 30° = angle of elevation of a lamp post from a point 10 m back from the road

In Δ ACB, we have

θ = 45°

Opposite side = AB

Adjacent side = BC

∴ $tan \:45 = \frac{AB}{BC}$

$\implies 1 = \frac{AB }{BC}$

$\implies AB = BC$ ...... (i)

In Δ ADB, we have

θ = 30°

Opposite side = AB

Adjacent side = BD = BC + CD = BC + 10

∴ $tan \:30 = \frac{AB}{BD}$

$\implies \frac{1}{\sqrt{3}} = \frac{AB }{BC + 10}$

$\implies BC + 10 = AB \sqrt{3}$

substituting from (i)

$\implies BC + 10 = BC\sqrt{3}$

$\implies BC\sqrt{3} - BC = 10$

$\implies BC (\sqrt{3} - 1 ) = 10$

$\implies BC = \frac{10}{\sqrt{3} - 1}$

$\implies BC = \frac{10}{1.732 - 1}$

$\implies BC = \frac{10}{0.732}$

$\implies \bold{BC = 13.66\: m}$

Thus, the width of the road is 13.66 m.

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