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,
[tex]tan(20^{\circ})=\frac{AB}{x}[/tex]
[tex]tan(20^{\circ})x=AB[/tex]
[tex]0.364x=AB[/tex] ..... (1)
In triangle ABM,
[tex]tan(50^{\circ})=\frac{AB}{7-x}[/tex]
[tex]1.192(7-x)=AB[/tex] ..... (2)
From (1) and (2), we get
[tex]0.364x=1.192(7-x)[/tex]
[tex]1.556x=8342[/tex]
[tex]x=5.36[/tex]
The length of AB is,
[tex]AB=0.364(5.36)=1.95147\approx 1.95[/tex]
Therefore, the he bird is at a height of 1.95 km from the ground.
