Question

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of
depression from the top of the tower of a foot of the hill is 30°. If the tower is 50 m high, find
the height of the hill​

Answer 1

Given :-

• AB is the hill having h metre height

• DC the tower having 50 meter height.

• Angle ACB is 60° and Angle DBC is 30° .

• The height of hill H = 50 + y m

Solution :-

Taking the triangle DBC .

Here Perpendicular is 50m ( DC ) and Base is x m(CB)

Angle DCB is 90° and angle DBC is 30° .

$\sf{ \red{ : \implies} \: \tan( \theta) \: = \dfrac{perpendicular}{base} }$

$\sf{ \red{ : \implies} \: \tan( 30) \: = \dfrac{50}{x} }$

$\sf{ \red{ : \implies} \: \dfrac{1}{\sqrt{3}} \: = \dfrac{50}{x} }$

$\sf{ \red{ : \implies} \: 50 \sqrt{3} \: = x }$

Now base of triangle DBC is equivalent to the base of triangle ABC.

Now taking triangle ABC

Here Perpendicular is AB( H meter ) and base is CB(50√3 m)

Angle ABC is 90° and Angle ACB is 60°

$\sf{ \red{ : \implies} \: \tan( \theta) \: = \dfrac{perpendicular}{base} }$

$\sf{ \red{ : \implies} \: \tan( 60) \: = \dfrac{H}{50 \sqrt{3} } }$

$\sf{ \red{ : \implies} \: \sqrt{3} \: = \dfrac{H}{50 \sqrt{3} } }$

$\sf{ \red{ : \implies} \: \sqrt{3} \times 50 .\sqrt{3} = H }$

$\sf{ \red{ : \implies} \: 3 \times 50 = H }$

• $\underline{\underline{\sf{ \red{ : \implies} \: 150 = H }}}$

Answer 2

Answer:

Given :-

AB is the hill having h metre height

DC the tower having 50 meter height.

Angle ACB is 60° and Angle DBC is 30° .

The height of hill H = 50 + y m

Solution :-

Taking the triangle DBC .

Here Perpendicular is 50m ( DC ) and Base is x m(CB)

Angle DCB is 90° and angle DBC is 30° .

\sf{ \red{ : \implies} \: \tan( \theta) \: = \dfrac{perpendicular}{base} }:⟹tan(θ)=

base

perpendicular

\sf{ \red{ : \implies} \: \tan( 30) \: = \dfrac{50}{x} }:⟹tan(30)=

x

50

\sf{ \red{ : \implies} \: \dfrac{1}{\sqrt{3}} \: = \dfrac{50}{x} }:⟹

3

1

=

x

50

\sf{ \red{ : \implies} \: 50 \sqrt{3} \: = x }:⟹50

3

=x

Now base of triangle DBC is equivalent to the base of triangle ABC.

Now taking triangle ABC

Here Perpendicular is AB( H meter ) and base is CB(50√3 m)

Angle ABC is 90° and Angle ACB is 60°

\sf{ \red{ : \implies} \: \tan( \theta) \: = \dfrac{perpendicular}{base} }:⟹tan(θ)=

base

perpendicular

\sf{ \red{ : \implies} \: \tan( 60) \: = \dfrac{H}{50 \sqrt{3} } }:⟹tan(60)=

50

3

H

\sf{ \red{ : \implies} \: \sqrt{3} \: = \dfrac{H}{50 \sqrt{3} } }:⟹

3

=

50

3

H

\sf{ \red{ : \implies} \: \sqrt{3} \times 50 .\sqrt{3} = H }:⟹

3

×50.

3

=H

\sf{ \red{ : \implies} \: 3 \times 50 = H }:⟹3×50=H

\underline{\underline{\sf{ \red{ : \implies} \: 150 = H }}}

:⟹150=H

Step-by-step explanation:

$\huge\boxed{\fcolorbox{blue}{orange}{hope it helps}}$