QUESTION IMAGE
Question
solve the following triangle. b = 30°, c = 70°, b = 5 a≈ □° (simplify your answer. a≈ □ (type an integer or decimal rounded to two decimal places as needed.) c≈ □ (type an integer or decimal rounded to two decimal places as needed.)
Step1: Find angle \( A \)
Use the angle - sum property of a triangle \( A + B + C=180^{\circ} \).
Given \( B = 30^{\circ} \) and \( C = 70^{\circ} \), then \( A=180^{\circ}-(B + C)=180^{\circ}-(30^{\circ}+70^{\circ}) = 80^{\circ} \).
Step2: Use the sine rule \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
First, find \( a \). Since \(\frac{a}{\sin A}=\frac{b}{\sin B}\), and \( b = 5\), \(A = 80^{\circ}\), \(B = 30^{\circ}\).
We know that \(\sin A=\sin80^{\circ}\approx0.9848\), \(\sin B=\sin30^{\circ}=0.5\).
From \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we can solve for \( a\): \(a=\frac{b\sin A}{\sin B}\).
Substitute the values: \(a=\frac{5\times\sin80^{\circ}}{\sin30^{\circ}}=\frac{5\times0.9848}{0.5}=9.848\approx9.85\).
Step3: Find \( c \)
Using the sine rule \(\frac{c}{\sin C}=\frac{b}{\sin B}\), \(\sin C=\sin70^{\circ}\approx0.9397\), \(\sin B = 0.5\), \(b = 5\).
Then \(c=\frac{b\sin C}{\sin B}=\frac{5\times\sin70^{\circ}}{\sin30^{\circ}}=\frac{5\times0.9397}{0.5}=9.397\approx9.40\).
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\(A\approx80^{\circ}\), \(a\approx9.85\), \(c\approx9.40\)