1.

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the pedestal is 45'. Find the height of the pedestal.

Answer»

Correct Question :–

A statue 1.6 m tall, stands on the TOP of the pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the HEIGHT of the pedestal.

Answer :–

The height of the pedestal = \boxed{ \tt \orange{h = 0.8( \sqrt{3} + 1) m}}

To Find :–

  • Height of the pedestal.

Solution :–

Let the height of pedestal BD be h.

  • ∠C = 60°
  • ∠D = 45°
  • CD = 1.6cm

i.e. ∠CAB = 60°, ∠DAB = 45° and CD = 1.6cm

In the right angled TRIANGLE ABD, we have,

→ \tan(45 \degree) = \dfrac{BD}{AB}

We know that, tan 45° is 1.

→ 1 = \dfrac{h}{AB}

→ AB = h.............(1)

In right angled triangle ABC, we have,

→ \tan(60 \degree) = \dfrac{BC}{AB}

We know that, tan 60° is √3

→ \sqrt{3} = \dfrac{BD + DC}{AB}

→ \sqrt{3} = \dfrac{h + 1.6}{AB}

→ ab = \dfrac{h + 1.6}{ \sqrt{3} } .............(2)

Now,

Comparing the both EQUATIONS (1) & (2), we GET,

→ h = \dfrac{h + 1.6}{ \sqrt{3} }

→ \sqrt{3} h = h + 1.6

→ \sqrt{3} h - h = 1.6

→ h( \sqrt{3} - 1) = 1.6

→ h = \dfrac{1.6}{ \sqrt{3} - 1 } \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1}

→ h = \dfrac{1.6( \sqrt{3} + 1) }{ {( \sqrt{3}) }^{2} - {(1)}^{2} }

→ h = \dfrac{1.6( \sqrt{3} + 1) }{3 - 1}

→ h = \dfrac{ \cancel{1.6}( \sqrt{3} + 1) }{ \cancel{2} }

→ h = \dfrac{0.8( \sqrt{3} +1 )}{1}

→ \boxed{ \tt \green{h = 0.8( \sqrt{3} + 1)}}

Hence,

The height of the pedestal = \boxed{ \tt \red{h = 0.8( \sqrt{3} + 1) m}}


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