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Question
christopher is solving the triangle by finding all the missing sides and angles.
part 1:
$x^{2}=8^{2}+14^{2}-2(8)(14)cos 67$
$x^{2}=64 + 196 - 224cos 67$
$x^{2}=172.5$
$x = 13.1$
part 2: (angle opposite 8)
$\frac{\sin 67}{13.1}=\frac{\sin \theta}{8}$
$\sin 67\cdot 8 = 13.1\sin \theta$
$12.9 = 13.1\sin \theta$
$0.9837 = \sin \theta$
$\theta = 76.6$
part 3: (angle opposite 14)
$\frac{\sin 67}{13.1}=\frac{\sin \theta}{14}$
$\sin 67\cdot 13.1 = 14\sin \theta$
$12.1 = 14\sin \theta$
$0.8613 = \sin \theta$
$\theta = 59.5$
Step1: Check the Law of Cosines application
The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Here, \(a = 8\), \(b=14\), \(C = 67^{\circ}\), and \(c=x\).
Using \(\cos67^{\circ}\approx0.3907\), we have \(224\times0.3907 = 87.5\), then \(x^{2}=64 + 196-87.5=172.5\), and \(x=\sqrt{172.5}\approx13.1\) (correct).
Step2: Check the Law of Sines application for the angle opposite \(8\)
The Law of Sines formula is \(\frac{\sin A}{a}=\frac{\sin B}{b}\). Let \(A = 67^{\circ}\), \(a = 13.1\), \(b = 8\), and the angle opposite \(8\) be \(\theta\).
Since \(\sin67^{\circ}\approx0.9205\), then \(\sin\theta=\frac{8\times0.9205}{13.1}=\frac{7.364}{13.1}\approx0.5622\), and \(\theta=\sin^{- 1}(0.5622)\approx34.2^{\circ}
eq76.6^{\circ}\) (error).
Step3: Check the Law of Sines application for the angle opposite \(14\)
Using the Law of Sines \(\frac{\sin A}{a}=\frac{\sin B}{b}\), with \(A = 67^{\circ}\), \(a = 13.1\), \(b = 14\)
Since \(\sin67^{\circ}\approx0.9205\), then \(\sin\theta=\frac{14\times0.9205}{13.1}=\frac{12.887}{13.1}\approx0.9837\), and \(\theta=\sin^{-1}(0.9837)\approx79.6^{\circ}
eq59.5^{\circ}\) (error).
The error in Part 2: When using the Law of Sines \(\frac{\sin67^{\circ}}{13.1}=\frac{\sin\theta}{8}\), \(\sin\theta=\frac{8\times\sin67^{\circ}}{13.1}\approx\frac{8\times0.9205}{13.1}\approx0.5622\), \(\theta\approx34.2^{\circ}\).
The error in Part 3: After finding one - correct angle (say we use the Law of Cosines or correct Law of Sines), using the angle - sum property of a triangle (\(A + B + C=180^{\circ}\)). If \(C = 67^{\circ}\), and we find one non - error angle (from corrected Part 2 \(\theta\approx34.2^{\circ}\)), then the third angle is \(180-(67 + 34.2)=78.8^{\circ}\)
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Part 1 calculation of \(x = 13.1\) is correct. Part 2 has an error: the angle opposite \(8\) is approximately \(34.2^{\circ}\) (not \(76.6^{\circ}\)). Part 3 has an error: using the angle - sum property of a triangle (after correcting Part 2), the angle opposite \(14\) is approximately \(78.8^{\circ}\) (not \(59.5^{\circ}\))