Question

solve the right triangle using the given information
a = 77.5 yd, b = 41.8 yd

c = \square yd
(simplify your answer. type an integer or a decimal. round to the nearest tenth if needed.)
a = \square^{\circ}\square
(simplify your answers. type integers. round to the nearest ten minutes if needed.)
b = \square^{\circ}\square
(simplify your answers. type integers. round to the nearest ten minutes if needed.)

Explanation:

Step1: Find hypotenuse \( c \) using Pythagorean theorem

The Pythagorean theorem for a right triangle is \( c = \sqrt{a^2 + b^2} \). Given \( a = 77.5 \) yd and \( b = 41.8 \) yd.
First, calculate \( a^2 \) and \( b^2 \):
\( a^2 = (77.5)^2 = 6006.25 \)
\( b^2 = (41.8)^2 = 1747.24 \)
Then, \( a^2 + b^2 = 6006.25 + 1747.24 = 7753.49 \)
So, \( c = \sqrt{7753.49} \approx 88.0 \) yd (rounded to the nearest tenth).

Step2: Find angle \( A \) using tangent function

\( \tan(A) = \frac{a}{b} \) (since \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) for angle \( \theta \)).
\( \tan(A) = \frac{77.5}{41.8} \approx 1.8541 \)
Then, \( A = \arctan(1.8541) \approx 61.7^\circ \). Convert the decimal part to minutes: \( 0.7^\circ \times 60 = 42' \), so \( A \approx 61^\circ 40' \) (rounded to nearest ten minutes).

Step3: Find angle \( B \) using tangent function

\( \tan(B) = \frac{b}{a} \)
\( \tan(B) = \frac{41.8}{77.5} \approx 0.5394 \)
Then, \( B = \arctan(0.5394) \approx 28.3^\circ \). Convert the decimal part to minutes: \( 0.3^\circ \times 60 = 18' \), so \( B \approx 28^\circ 20' \) (rounded to nearest ten minutes). Also, since \( A + B = 90^\circ \), we can check: \( 61^\circ 40' + 28^\circ 20' = 90^\circ \), which is correct.

Answer:

\( c = \boxed{88.0} \) yd
\( A = \boxed{61^\circ 40'} \)
\( B = \boxed{28^\circ 20'} \)