Question

Susan and Mark are standing at different places on a beach and watching a bird. The angles of elevation they make are 20º and 50º, respectively. If Susan and Mark are 7 kilometers apart and the bird is between them, the bird is at a height of kilometers from the ground. NextReset

Answer 1

Answer:

The bird is at a height of 1.95 km from the ground.

Step-by-step explanation:

It is given that the distance between susan and mark is 7 km and bird is between  susan and mark. The angles of elevation they make are 20º and 50º, respectively.

Draw a perpendicular line from the bird on base.

Let the distance of susan from and the altitude be x.

In triangle ABS,

$tan(20^{\circ})=\frac{AB}{x}$

$tan(20^{\circ})x=AB$

$0.364x=AB$       ..... (1)

In triangle ABM,

$tan(50^{\circ})=\frac{AB}{7-x}$

$1.192(7-x)=AB$     ..... (2)

From (1) and (2), we get

$0.364x=1.192(7-x)$

$1.556x=8342$

$x=5.36$

The length of AB is,

$AB=0.364(5.36)=1.95147\approx 1.95$

Therefore, the he bird is at a height of 1.95 km from the ground.

Answer 2

The answer is 1.952 km