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solve for the remaining angles and side of the two triangles that can b…

Question

solve for the remaining angles and side of the two triangles that can be created. round to the nearest hundredth: a = 60°, a = 7, b = 8

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
Substitute \(A = 60^{\circ}\), \(a = 7\), and \(b = 8\) into the formula: \(\frac{7}{\sin60^{\circ}}=\frac{8}{\sin B}\).
Cross - multiply to get \(7\sin B=8\sin60^{\circ}\).
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\), then \(7\sin B = 8\times0.866 = 6.928\).
So, \(\sin B=\frac{6.928}{7}\approx0.9897\).

Step2: Find angle \(B\)

Since \(\sin B\approx0.9897\), then \(B=\sin^{- 1}(0.9897)\approx81.11^{\circ}\) or \(B = 180^{\circ}-81.11^{\circ}=98.89^{\circ}\) (because \(\sin\theta=\sin(180^{\circ}-\theta)\)).

Step3: Find angle \(C\) for the first case (\(B = 81.11^{\circ}\))

Using the angle - sum property of a triangle (\(A + B + C=180^{\circ}\)), when \(A = 60^{\circ}\) and \(B = 81.11^{\circ}\), then \(C=180^{\circ}-(60^{\circ}+81.11^{\circ})=38.89^{\circ}\).

Step4: Find side \(c\) for the first case (\(B = 81.11^{\circ}\))

Using the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\).
Substitute \(a = 7\), \(A = 60^{\circ}\), and \(C = 38.89^{\circ}\) (where \(\sin38.89^{\circ}\approx0.628\) and \(\sin60^{\circ}\approx0.866\)).
\(c=\frac{7\times\sin38.89^{\circ}}{\sin60^{\circ}}=\frac{7\times0.628}{0.866}\approx5.07\).

Step5: Find angle \(C\) for the second case (\(B = 98.89^{\circ}\))

Using the angle - sum property of a triangle (\(A + B + C = 180^{\circ}\)), when \(A=60^{\circ}\) and \(B = 98.89^{\circ}\), then \(C=180^{\circ}-(60^{\circ}+98.89^{\circ})=21.11^{\circ}\).

Step6: Find side \(c\) for the second case (\(B = 98.89^{\circ}\))

Using the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\).
Substitute \(a = 7\), \(A = 60^{\circ}\), and \(C = 21.11^{\circ}\) (where \(\sin21.11^{\circ}\approx0.360\) and \(\sin60^{\circ}\approx0.866\)).
\(c=\frac{7\times\sin21.11^{\circ}}{\sin60^{\circ}}=\frac{7\times0.360}{0.866}\approx2.91\).

Answer:

First triangle:

  • \(B\approx81.11^{\circ}\), \(C\approx38.89^{\circ}\), \(c\approx5.07\)

Second triangle:

  • \(B\approx98.89^{\circ}\), \(C\approx21.11^{\circ}\), \(c\approx2.91\)