Question

A helicopter flies between two islands that are 20 miles apart. The angle of elevation with one island is 15o and 35o with the second island. What is the mileage of the helicopter? Round your answer to the nearest tenth of a mile. (Law of Sines or Law of Cosines Problem)

Answer 1

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) = $\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{h}{x}$

h = x.tan(15) ---------(1)

From ΔACD,

tan(35) = $\frac{h}{12-x}$

h = (12 - x).tan(35) ------(2)

By equating the values of h from equations (1) and (2),

x.tan(15) = (12 - x).tan(35)

$\frac{x}{12-x}=\frac{\text{tan}35}{\text{tan}15}$

$\frac{x}{12-x}=2.613$

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.