Question

question 8 (mandatory) (1 point)
determine the measure of the angle \\( \theta \\) to the nearest degree.

a) \\( \theta=36^{\circ} \\)
b) \\( \theta=41^{\circ} \\)
c) \\( \theta=39^{\circ} \\)
d) \\( \theta=47^{\circ} \\)

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines states that \(c^{2}=a^{2}+b^{2}-2ab\cos C\). First, find the length of side \(AC\) (let's call it \(a\)) using the Law of Cosines:
\(a^{2}=11.1^{2}+10.0^{2}-2\times11.1\times10.0\times\cos41^{\circ}\)
\(a^{2}=123.21 + 100-222\times0.7547\)
\(a^{2}=223.21-167.5434\)
\(a^{2}=55.6666\)
\(a=\sqrt{55.6666}\approx7.46\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{\sin\theta}{10.0}=\frac{\sin41^{\circ}}{7.46}\)
\(\sin\theta=\frac{10.0\times\sin41^{\circ}}{7.46}\)
\(\sin\theta=\frac{10.0\times0.6561}{7.46}\)
\(\sin\theta=\frac{6.561}{7.46}\approx0.88\)
\(\theta=\sin^{- 1}(0.88)\approx61.6^{\circ}\) (Wait, no, wrong approach. Let's start over correctly)

Correct approach:
Using the Law of Cosines formula for \(\theta\) (angle \(A\)):
\(\cos\theta=\frac{b^{2}+c^{2}-a^{2}}{2bc}\) where \(a = BC = 10.0\), \(b = AC\), \(c = AB=11.1\). Wait, no, better use Law of Sines directly.
By the Law of Sines: \(\frac{\sin\theta}{BC}=\frac{\sin C}{AB}\)
\(\sin\theta=\frac{BC\times\sin C}{AB}\)
Substitute \(BC = 10.0\), \(C = 41^{\circ}\), \(AB = 11.1\)
\(\sin\theta=\frac{10.0\times\sin41^{\circ}}{11.1}\)
\(\sin\theta=\frac{10.0\times0.6561}{11.1}\)
\(\sin\theta=\frac{6.561}{11.1}\approx0.591\)
\(\theta=\sin^{-1}(0.591)\approx36.2^{\circ}\approx36^{\circ}\)

Answer:

A. \(\theta = 36^{\circ}\)