Question

please solve this it's an emergency​

Answer 1

height of flag staff is 5m

Angle 1 is 45

Angle 2 is 60

let the height of the tower is x

in triangle with the bottom of flagstaff tan45=h/x

h=x

in triangle with the top of the flagstaff is tan60=5+h/x

√3x=5+h

since x=h

√3x-x=5

x(√3-1)=5

x=5/√3-1=h

Answer 2

Answer:

Height of flag staff (h) = 5m

Let height of tower = y m

Let BC = x m

In triangle DBC

$\frac{y}{x} = tan45 \\ \\ \frac{y}{x} = 1 \\ \\ y = x \: \: \: \: \: \: \: \: - - eq(1)$

Now in triangle ABC

$\frac{5 + y}{x} = tan60 \\ \\ \frac{5 + x}{x} = \sqrt{3} \\ \\ 5 + x = \sqrt{3} \: x \\ \sqrt{3} \: x - x = 5 \\ x( \sqrt{3} - 1) = 5 \\ \\ x = \frac{5}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1} \\ \\ x = \frac{5 \sqrt{3} + 5 }{3 - 1} \\ \\ x =\frac{5 \sqrt{3} + 5 }{2} m$

By eq(1)

$y = x \\ \\ y =\frac{5 \sqrt{3} + 5 }{2} m$

Therefore height of tower is

$y = \frac{5 \sqrt{3} + 5 }{2} m$

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