A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30° and 60° respectively. What is the distance the two cars and how far is each car from the tower?
Solution
Let AB be the height of tower 50 m and angle of depression from the top of the tower are 60° and 30° respectively at two observing Car C and D.
Let BD = x m, CD = y m and ∠ADB = 30°, ∠ACb = 60°
We have the corresponding figure as follows
So we use trigonometric ratios.
In a triangle ABD
`=> tan D = (AB)/(BD)`
`=> tan 30° = 50/x`
`=> 1/sqrt3 = 50/x`
`=> x = 50sqrt3`
Since x = 86.6
Again in a triangle ABC
`=> tan C = (AB)/(BC)`
`=> tan 60° = 50/(x - y)`
`=> sqrt3 = 50/(x - y)`
`=> sqrtt3 xx 50sqrt3 - sqrt3y = 50`
`=> y = 57.67`
Therefore x - y= 86.6 - 57.67
`=> x - y = 28.93`
Hence the distance of first car from tower is 86.6 m
And the distance of second car from tower is 57.67 m
And the distance between cars is 28.93 m
