Question

Two men are on the opposite sides of a tower. The measure of the top of the angle is 30 and 60. Find the distance between them if the height of the tower is 80m

Answer 1

let BD be the required distance between two men .

Given, height of the tower = 80 cm ,

so now we have

In ∆ ACB ,
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tan60° = AC/ BC

√3 = 80 / BC

BC = 80 /√3

BC = (80/√3 ) × ( √3 /√3 )

BC = 80√3/ 3 cm

from ∆ ADC ,
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tan30° = AC / CD

1 /√3 = 80/ CD

CD = 80√3 cm

distance between two men = BD

BD = BC + CD

BD = ( 80√3/ 3 ) + 80√3

BD = (80√3 + 240√3 ) /3

BD = (320√3) / 3

BD = 106.67 √3 = 184.7 cm

Answer :the distance between two men

= 184.7 cm

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