Answer to Question #146983 in Trigonometry for niema

Question

Two hot-air balloons are moored to level ground below, each at a different location. An observer at each location determines the angle of elevation to the opposite balloon as shown at the right. The observers are 2.0 km apart.
a) What is the distance separating the balloons, to the nearest tenth of a kilometre?
b) Determine the difference in height (above the ground) between the two balloons. Round your answer to the nearest metre.

Answer 1

(1)

In the above diagram and triangle ABC

We find the side through sin formula

Sin(100)⁰ × 2 = AC

AC = 1.96Km

and AB = Sin(25)⁰ × 1.96

AB = 0.80 Km

In the above diagram and triangle BCD

We find the side through sin formula

Sin(105)⁰ × 2 = BD

BD = 1.93Km

and CD = sin(20)⁰ × 1.93

CD = 0.62 Km

Now from Diagram and triangle ABD

We find side AD through two 2 sides

AB and BD

(AD)² = (AB)² + (BD)² + AB×BD×cos (80) ⁰

AD² = (0.80)² + (1.93) + 0.80×1.93× 0.17

AD² = 4.72637

AD = 2.15 Km

the distance separating the balloons, to the nearest tenth of a kilometre is 2.15 Km

(2)

difference in height (above the ground) between the two balloons. Round your answer to the nearest metre.

AB - CD

AB = 0.80 Km = 800m

CD = 0.62 Km = 620m

AB - CD = 800 - 620

= 180 m .