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 \( 90^{\circ} \).
If one non - right angle is \( 49^{\circ} \), then \( A=90^{\circ}-49^{\circ}=41^{\circ} \)

Step2: Find side \( c \)

We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here, \(\theta = 49^{\circ}\), adjacent side to \(49^{\circ}\) is \(18\), and hypotenuse is \(c\).
\(\cos49^{\circ}=\frac{18}{c}\), so \(c=\frac{18}{\cos49^{\circ}}\). Using a calculator, \(\cos49^{\circ}\approx0.656\), then \(c=\frac{18}{0.656}\approx27.4\)

Step3: Find side \( b \)

We know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 49^{\circ}\), opposite side to \(49^{\circ}\) is \(b\), and hypotenuse is \(c\approx27.4\) (we can also use \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\tan49^{\circ}=\frac{b}{18}\))
Using \(\tan49^{\circ}\approx1.150\), then \(b = 18\times\tan49^{\circ}\approx18\times1.150 = 20.7\)

Answer:

\(c\approx27.4\), \(b\approx20.7\), \(A = 41^{\circ}\)