Question
Solve the following exercises using the law of cosines and the law of sines In a triangle we have the lengths a)80 and b)50 and the angle C)50. What is the length of c and the angle of the resulting vector?
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Answer to a math question Solve the following exercises using the law of cosines and the law of sines In a triangle we have the lengths a)80 and b)50 and the angle C)50. What is the length of c and the angle of the resulting vector?
Jayne
4.496 Answers
1. **Apply the Law of Cosines:**
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
c^2 = 80^2 + 50^2 - 2 \cdot 80 \cdot 50 \cdot \cos(50^\circ)
c^2 = 6400 + 2500 - 8000 \cdot \cos(50^\circ)
c^2 = 8900 - 5146.67
c^2 = 3753.33
c\approx61.30
2. **Use the Law of Sines to find \(\angle B\):**
\frac{b}{\sin(\angle B)} = \frac{c}{\sin(\angle C)}
\frac{50}{\sin(\angle B)}=\frac{61.30}{\sin(50^{\circ})}
\sin(\angle B)=\frac{50\cdot\sin(50^{\circ})}{61.30}
\sin(\angle B)\approx0.999
\angle B\approx\sin^{-1}(0.999)
\angle B\approx88.67^{\circ}
Therefore, the length of side \(c\) is approximately \(61.3\) and the angle \(\angle B\) is approximately\(88.67^\circ\)
2. **Use the Law of Sines to find \(\angle B\):**
Therefore, the length of side \(c\) is approximately \(61.3\) and the angle \(\angle B\) is approximately
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