Question

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

20($\sqrt{3} -1$) metre

Step-by-step explanation:

let C be the top of the tower CA and A be the top of the building AB and P be the point on ground.

in triangle ABP

$tan 45 =\frac{AB}{BP}$

$1 = \frac{20}{BP}$

BP = 20 m                            ....(1)

in triangle PBC

$tan 60 =\frac{AB}{BP}$

$\sqrt{3} =\frac{AC+AB}{BP}$

BP $\sqrt{3}$  =  AC+AB

20 $\sqrt{3}$ =  AC+20                  [USING...(1) ]

20 $\sqrt{3}$ - 20 = AC

20( $\sqrt{3}$ -1) =AC

20 ( 1.73 - 1 ) =AC

20 × 0.73 = AC

14.60 m

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