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 the foot of the hill is 30°. If the tower is 50 metre high, find the height of the hill.​

Answer 1

Answer:

Let AB be the tower and CD be the hill. Then, ∠ACB=30o,∠CAD=60o and AB=50 m.

Let CD=x m

In right △BAC, we have,

cot30o=ABAC

3=50AC

AC=503 m

In right △ACD, we have,

tan60o=ACCD

3=503x

x=50×3=150 m

Therefore, the height of the hill is 150 m.

Answer 2

$\large\underline{\bold{Solution-}}$

Let AB be the tower and CD be the hill.

So, 

• ∠ACB = 30°

• ∠CAD=60° 

and 

• height of tower, AB = 50 m.

• Let height of the hill, CD = 'x' meter.

Now,

• In right △BAC,

we have,

$\sf \: tan30 \degree \: = \: \dfrac{AB}{AC}$

$\sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{50}{AC}$

$\therefore \bf \: AC \: = \: 50 \sqrt{3} \: m$

Now,

• In right △ACD,

we have,

$\sf \: tan60 \degree \: = \: \dfrac{CD}{AC}$

$\sf \: \sqrt{3} = \dfrac{x}{50 \sqrt{3} } \: \: \: \: \: \: \: ( \because \: AC \: = \: 50 \sqrt{3} )$

$\therefore \bf\: x = 50 \sqrt{3} \times \sqrt{ 3} = 150 \: m.$

Therefore, the height of the hill, CD is 150 m.

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Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side