Question

solve the right triangle.
round your answers to the nearest tenth.

Explanation:

Step1: Find angle \( A \)

In a right - triangle, the sum of the two non - right angles is \( 90^{\circ} \). Given one non - right angle is \( 49^{\circ} \), then \( A=90^{\circ}-49^{\circ}=41^{\circ} \)

Step2: Find side \( c \)

Using the cosine function \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). Here \( \theta = 49^{\circ} \), adjacent side to \( 49^{\circ} \) is \( 18 \), and hypotenuse is \( c \). So \( \cos49^{\circ}=\frac{18}{c} \), then \( c=\frac{18}{\cos49^{\circ}} \). Since \( \cos49^{\circ}\approx0.656 \), \( c=\frac{18}{0.656}\approx27.4 \)

Step3: Find side \( b \)

Using the sine function \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \). Here \( \theta = 49^{\circ} \), opposite side to \( 49^{\circ} \) is \( b \), and hypotenuse is \( c\approx27.4 \). So \( \sin49^{\circ}=\frac{b}{c} \), substituting \( c\approx27.4 \), \( b = c\times\sin49^{\circ}\approx27.4\times0.755\approx20.7 \)

Answer:

\( c = 27.4 \), \( b = 20.7 \), \( A = 41^{\circ} \)