Answer:
Step-by-step explanation:
From the figure attached,
A helicopter is flying at a height 'h' and angle of elevations from two islands C and B are 35° and 15° respectively.
Let the measure of BD = x miles
From ΔABD,
tan(15) = [tex]\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{h}{x}[/tex]
h = x.tan(15) ---------(1)
From ΔACD,
tan(35) = [tex]\frac{h}{12-x}[/tex]
h = (12 - x).tan(35) ------(2)
By equating the values of h from equations (1) and (2),
x.tan(15) = (12 - x).tan(35)
[tex]\frac{x}{12-x}=\frac{\text{tan}35}{\text{tan}15}[/tex]
[tex]\frac{x}{12-x}=2.613[/tex]
x = 2.613(12 - x)
x + 2.613x = 31.386
3.613x = 31.386
x = 9.914 ≈ 9.9 miles
From equation (1),
h = x.tan(15)
= 9.9[tan(15)]
= 2.65 ≈ 2.7 miles
Therefore, helicopter is flying at the vertical height = 2.7 miles and horizontal distance of the helicopter from the island B = 9.9 miles.
