QUESTION IMAGE
Question
find all missing elements.
round to the nearest tenth.
Step1: Use the Law of Cosines to find angle \(A\)
The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos A\).
Substitute \(a = 10\), \(b = 7\), \(c = 6\) into the formula:
\(10^{2}=7^{2}+6^{2}-2\times7\times6\times\cos A\)
\(100=49 + 36-84\cos A\)
\(100=85-84\cos A\)
\(84\cos A=85 - 100=-15\)
\(\cos A=\frac{-15}{84}\approx - 0.1786\)
\(A=\cos^{-1}(-0.1786)\approx100.3^{\circ}\)
Step2: Use the Law of Sines to find angle \(B\)
The Law of Sines formula is \(\frac{\sin B}{b}=\frac{\sin A}{a}\)
We know \(a = 10\), \(b = 7\), \(A\approx100.3^{\circ}\)
\(\sin B=\frac{b\sin A}{a}=\frac{7\times\sin(100.3^{\circ})}{10}\)
\(\sin(100.3^{\circ})\approx0.984\)
\(\sin B=\frac{7\times0.984}{10}=0.6888\)
\(B=\sin^{-1}(0.6888)\approx43.5^{\circ}\)
Step3: Use the angle - sum property of a triangle to find angle \(C\)
Since \(A + B + C=180^{\circ}\)
\(C=180^{\circ}-A - B\)
\(C=180^{\circ}-100.3^{\circ}-43.5^{\circ}=36.2^{\circ}\)
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\(A\approx100.3^{\circ}\), \(B\approx43.5^{\circ}\), \(C\approx36.2^{\circ}\)