Question

solve the right triangle. round your answers to the nearest tenth. b = \\(\square^\circ\\) a = \\(\square\\) c = \\(\square\\) 5 ×

Explanation:

Step1: Find angle \( B \)

In a right - triangle, the sum of the two non - right angles is \( 90^{\circ} \).
If one angle \( A = 50^{\circ} \), then \( B=90^{\circ}-A \).
\( B = 90^{\circ}-50^{\circ}=40^{\circ} \)

Step2: Find side \( a \)

We know that \(\tan A=\frac{a}{23}\) (since \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\) in a right - triangle, and \( A = 50^{\circ}\), adjacent side \( = 23\), opposite side \(=a\)).
\(a = 23\times\tan(50^{\circ})\)
Using a calculator, \(\tan(50^{\circ})\approx1.191754\)
\(a\approx23\times1.191754 = 27.4\)

Step3: Find side \( c \)

We know that \(\cos A=\frac{23}{c}\) (since \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) in a right - triangle, \(A = 50^{\circ}\), adjacent side \( = 23\), hypotenuse \(=c\)).
\(c=\frac{23}{\cos(50^{\circ})}\)
Using a calculator, \(\cos(50^{\circ})\approx0.642788\)
\(c=\frac{23}{0.642788}\approx35.8\)

Answer:

\(B = 40^{\circ}\), \(a\approx27.4\), \(c\approx35.8\)