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solve for the remaining angle and sides of the triangle described below…

Question

solve for the remaining angle and sides of the triangle described below. round to the nearest hundred

$b = 50^{circ},c = 10^{circ},b = 4$

Explanation:

Step1: Find angle \(A\)

In a triangle, \(A + B + C=180^{\circ}\). Given \(B = 50^{\circ}\) and \(C = 10^{\circ}\), then \(A=180^{\circ}-(B + C)\)
\(A=180^{\circ}-(50^{\circ}+ 10^{\circ})=120^{\circ}\)

Step2: Use the Law of Sines to find side \(c\)

The Law of Sines is \(\frac{b}{\sin B}=\frac{c}{\sin C}\). We know \(b = 4\), \(B = 50^{\circ}\), \(C = 10^{\circ}\). So \(c=\frac{b\sin C}{\sin B}\)
\(\sin50^{\circ}\approx0.7660\), \(\sin10^{\circ}\approx0.1736\)
\(c=\frac{4\times0.1736}{0.7660}=\frac{0.6944}{0.7660}\approx0.91\)

Step3: Use the Law of Sines to find side \(a\)

Using the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\), so \(a=\frac{b\sin A}{\sin B}\)
\(\sin120^{\circ}=\frac{\sqrt{3}}{2}\approx0.8660\)
\(a=\frac{4\times0.8660}{0.7660}=\frac{3.464}{0.7660}\approx4.52\)

Answer:

\(A = 120^{\circ}\), \(a\approx4.52\), \(c\approx0.91\)