Question

3. The angles of elevation to an airplane from two points A and B on level ground are 55o and 72o respectively. A and B are 2.2 miles apart and the plane is east of both points. Find the altitude of the plane.

Answer 1

Answer:

CD = 5.9 mi

Step-by-step explanation:

Given:

∠DAC = 55°

∠DBC = 72°

AB = 2.2 mi

∠ABD = 180 - ∠DBC = 180 - 72 = 108°

∠ADB = 180 - ∠BAD - ∠ABD = 180 -55 - 108 = 17°

Using Sine Rule:

$\frac{AB}{sin{ADB} } = \frac{BD}{sin{BAD}}$

$\frac{2.2}{sin{17}} = \frac{BD}{sin{55}}$

Solving for BD:

BD = 6.2 mi

From ΔBDC

Sin(72) = CD / BD

Sin(72) = CD / 6.2

Solving for CD:

CD = 6.2*sin(72) = 5.9 mi