Question

The angle of elevation of the top of a church tower from a point on level ground 500m away is 16% find the height of the tower

Answer 1

Answer:

Step-by-step explanation:

Let height of the tower be h and BQ =x

In ΔPQB

tan60∘=hx

⇒3–√=hx

⇒h=3–√x………(1)

In ΔPQA

tan30∘=h20+x

⇒13√=h20+x

⇒3–√h=20+x

⇒3×−−−√3–√x=20+x [Using equation (1)]

⇒3x−x=20

⇒2x=20

⇒x=10

From equation (1)

h=103–√=10×1.732=17.32 m

So, Height of the tower =17.32 m.

Distance of tower from point A =20+10=30 m

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