Question

solve the right triangle abc, where c = 90°.
a = 77.7 yd, b = 41.7 yd

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

Explanation:

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

The Pythagorean theorem for a right triangle is \( c = \sqrt{a^2 + b^2} \). Given \( a = 77.7 \) yd and \( b = 41.7 \) yd, we substitute these values:
\( c = \sqrt{(77.7)^2 + (41.7)^2} \)
First, calculate \( (77.7)^2 = 77.7 \times 77.7 = 6037.29 \) and \( (41.7)^2 = 41.7 \times 41.7 = 1738.89 \).
Then, add these results: \( 6037.29 + 1738.89 = 7776.18 \).
Now, take the square root: \( c = \sqrt{7776.18} \approx 88.2 \) yd.

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

The tangent of angle \( A \) is \( \tan(A) = \frac{a}{b} \). So, \( \tan(A) = \frac{77.7}{41.7} \approx 1.8633 \).
To find \( A \), take the arctangent: \( A = \arctan(1.8633) \approx 61.8^\circ \).
Convert the decimal part to minutes: \( 0.8^\circ \times 60 = 48' \), so \( A \approx 61^\circ 50' \) (rounded to nearest ten minutes: 48' is closer to 50' than 40').

Step3: Find angle \( B \) using angle sum property

In a right triangle, \( A + B + C = 180^\circ \), and \( C = 90^\circ \), so \( B = 90^\circ - A \).
\( B = 90^\circ - 61.8^\circ = 28.2^\circ \).
Convert the decimal part to minutes: \( 0.2^\circ \times 60 = 12' \), so \( B \approx 28^\circ 10' \) (rounded to nearest ten minutes: 12' is closer to 10' than 20').

Answer:

\( c \approx 88.2 \) yd
\( A \approx 61^\circ 50' \)
\( B \approx 28^\circ 10' \)