Question

charlie is watching hot air balloons. balloon a has risen at a 56° angle. balloon b has risen at an 81° angle. if the distance from balloon a to the ground is 1,200 feet, how far is balloon b from balloon a? round your answer to the nearest whole number.

Explanation:

Step1: Find distance from Charlie to Balloon A

Let \( d_A \) be the distance from Charlie to Balloon A. Using sine for the right triangle (opposite side = 1200 ft, angle \( x = 56^\circ \)):
\( \sin(56^\circ) = \frac{1200}{d_A} \)
\( d_A = \frac{1200}{\sin(56^\circ)} \approx \frac{1200}{0.8290} \approx 1447.5 \) ft.

Step2: Find distance from Charlie to Balloon B

Let \( d_B \) be the distance from Charlie to Balloon B. Using sine for the right triangle (opposite side = 1200 ft, angle \( y = 81^\circ \)):
\( \sin(81^\circ) = \frac{1200}{d_B} \)
\( d_B = \frac{1200}{\sin(81^\circ)} \approx \frac{1200}{0.9877} \approx 1214.9 \) ft.

Step3: Find angle at Charlie between balloons

The angle between the two lines of sight is \( 81^\circ - 56^\circ = 25^\circ \).

Step4: Apply Law of Cosines to find distance between balloons

Let \( D \) be the distance between Balloon A and Balloon B.
Law of Cosines: \( D^2 = d_A^2 + d_B^2 - 2 \cdot d_A \cdot d_B \cdot \cos(25^\circ) \)
Substitute values:
\( d_A \approx 1447.5 \), \( d_B \approx 1214.9 \), \( \cos(25^\circ) \approx 0.9063 \)
\( D^2 \approx (1447.5)^2 + (1214.9)^2 - 2 \cdot 1447.5 \cdot 1214.9 \cdot 0.9063 \)
Calculate each term:
\( (1447.5)^2 \approx 2095256.25 \)
\( (1214.9)^2 \approx 1475982.01 \)
\( 2 \cdot 1447.5 \cdot 1214.9 \cdot 0.9063 \approx 2 \cdot 1447.5 \cdot 1214.9 \cdot 0.9063 \approx 3134442.3 \)
\( D^2 \approx 2095256.25 + 1475982.01 - 3134442.3 \approx 436795.96 \)
\( D \approx \sqrt{436795.96} \approx 661 \) ft.

Answer:

661