Question

Hola!❤❤

Plz solve these questions!!​

Answer 1

Please see the attachment

Answer 2

Let height of the tower (CD)=h m

Distance from A to B = 20 m

Distance from B to C = x m

<DAC = 45°

<DBC = 60°

i ) From ∆DAC,

tanA = DC/AC

=> tan45° = h/(20+x)

=> 1 = h/(20+x)

=> 20+x = h

=> x = h - 20 ----( 1 )

ii ) From ∆DCB,

tanB = DC/BC

=>tan60° = h/x

=> √3 = h/x

=> x = h/√3 ----( 2 )

From (1) and (2), we get

h-20 = h/√3

=> √3(h-20) = h

=> √3h -20√3 = h

=> (√3-1)h =20√3

=> (√3- 1)h = 20√3

=> h = 20√3/(√3-1)

=> h=[20√3(√3+1)]/[(√3-1)(√3+1)]

=> h = [20(3+√3)]/[(√3)²-1²]

=> h = [20(3+1.732)]/(3-1)

=> h = 10(4.732)

=> h = 47.32 m

Therefore,

Height of the tower (h) = 47.32m

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