Question

solve the right triangle abc, where c = 90°. a = 77.2 yd, b = 41.7 yd c = □ yd (simplify your answer. type an integer or a decimal. round to the nearest tenth as needed.) a = □° □ (simplify your answers. type integers. round to the nearest ten minutes as needed.) b = □° □ (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 \( a = 77.2 \) yd and \( b = 41.7 \) yd, we substitute these values:
\( c = \sqrt{(77.2)^2 + (41.7)^2} \)
First, calculate \( (77.2)^2 = 77.2 \times 77.2 = 5959.84 \) and \( (41.7)^2 = 41.7 \times 41.7 = 1738.89 \). Then sum them: \( 5959.84 + 1738.89 = 7698.73 \). Now take the square root: \( c = \sqrt{7698.73} \approx 87.7 \) yd.

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

The tangent of angle \( A \) is \( \tan(A) = \frac{a}{b} \). So \( \tan(A) = \frac{77.2}{41.7} \approx 1.8513 \). Then \( A = \arctan(1.8513) \). Calculate this: \( A \approx 61.7^\circ \). Now convert the decimal part to minutes. \( 0.7^\circ \times 60 = 42' \), so \( A \approx 61^\circ 40' \) (rounded to nearest ten minutes: 42' is closer to 40' or 50'? 42 - 40 = 2, 50 - 42 = 8, so 40'). Wait, actually, let's recalculate more accurately. \( \arctan(77.2/41.7) \): \( 77.2 \div 41.7 \approx 1.8513 \). \( \arctan(1.8513) \approx 61.73^\circ \). \( 0.73^\circ \times 60 \approx 43.8' \), which rounds to 40' or 50'? Wait, the problem says "round to the nearest ten minutes". So 43.8' is closer to 40' (difference 3.8) or 50' (difference 6.2)? So 40'. So \( A \approx 61^\circ 40' \).

Step3: Find angle \( B \)

Since \( A + B = 90^\circ \) (in a right triangle), \( B = 90^\circ - A \). \( 90^\circ - 61.73^\circ = 28.27^\circ \). Convert to minutes: \( 0.27^\circ \times 60 \approx 16.2' \), which rounds to 20'? Wait, no: 28.27° is 28° + 0.27×60' = 28°16.2', which rounds to nearest ten minutes: 16.2' is closer to 20'? Wait, 16.2 - 10 = 6.2, 20 - 16.2 = 3.8? Wait, no, nearest ten minutes: 10', 20', 30'... So 16.2' is closer to 20' (difference 3.8) than 10' (difference 6.2). Wait, but let's check with tangent: \( \tan(B) = \frac{b}{a} = \frac{41.7}{77.2} \approx 0.5402 \). Then \( B = \arctan(0.5402) \approx 28.3^\circ \). \( 0.3^\circ \times 60 = 18' \), so 28°20' (since 18' is closer to 20' than 10'? Wait, 18 - 10 = 8, 20 - 18 = 2, so 20'). So \( B \approx 28^\circ 20' \).

Answer:

\( c \approx 87.7 \) yd
\( A \approx 61^\circ 40' \)
\( B \approx 28^\circ 20' \)