Question

From a point P on a level ground, the angle of elevation of the top tower is 30 degree. If the tower is 200 m high the distance of point P from the foot of the tower is??...​

Answer 1

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• angle of elevation of the top tower is 30°
• Height of tower = 200 m

$\Large{\underline{\mathfrak{\orange{\bf{Find}}}}}$

• distance of point P from the foot of the tower ?

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

Here, Right triangle be PQR,

Where,

• RQ (tower) = 200 m
• <RPQ = 30°

Let,

• PQ = x m

Using trigonometry Formula

★ tan θ = perpendicular/Base

So, we take

➥ tan 30° = 200/x

[ tan 30° = 1/√3 ]

➥ 1/√3 = 200/x

➥x = 200 × √3

[ √3 = 1.732 ]

➥ x = 200 * 1.732

➥ x = 346.400

➥ x = 346.4 m

$\Large{\underline{\underline{\mathfrak{\red{\bf{\:Hence}}}}}}$

• Distance of point P from the foot of the tower Q will be = 346.4 m

________________

Answer 2

Answer:

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