Question

in fig angleOAG=40°,the angle of elevation is​

Answer 1

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Answer 2

∠ADB = 60° , ∠ACB = 45° , CD= 40 m, BD= x m

Let AB = h m be the height of the tower.

In ∆ ABD

tan 60° = AB/BD = P /B

√3 = h /x

h = x √3

x = h/√3 ………………………... (1)

In ∆ ABC

tan 45° = AB/BC

tan 45° = AB/(BD+DC)

1 = AB/(x+40)

1= h /(x+40)

h = x + 40 ………………………. (2)

Put the value of x from eq 1 in eq 2.

h = x + 40

h = h/√3 +40

h - h/ √3 = 40

(√3h -h ) /√3 = 40

h (√3 -1) = 40√3

h = 40√3 /(√3 -1)

h = [40√3 /(√3 - 1)] x [(√3 +1)/(√3 +1)]

[By rationalising]

h = [120 + 40√3] / [(√3)² - 1²]

h = [120 + 40× 1.732] / 2

h = (120 + 69.28)/2 = 189.28/2 = 94.64 m

Hence, the height of the tower is 94.64 m.