Question

Can someone help me answer this, thanks

Answer 1

`Height = 173.8ft`

Explanation:

Given

`\theta = 40^o`

`d = 200ft` --- distance from building

`h=6ft`--- height of the building

Let the height of the building be H. So:

`\tan(\theta) = (Opposite)/(Adjacent)`

`\tan(40) = (H)/(200)`

Make H the subject

`H = 200 * \tan(40)`

`H = 167.82`

The required height s:

`Height = H + h`

`Height = 167.82 + 6`

`Height = 173.82ft`

`Height = 173.8ft` --- approximately