Question

find the measure of θ in the triangle. do not use a calculator.
b = □°

Explanation:

Step1: Identify the triangle sides

We have triangle \(ABC\) with \(AB = 4\), \(BC = 4\), and \(AC = 4\sqrt{3}\). We can use the Law of Cosines to find angle \(\theta\) (angle at \(B\)). The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(\theta\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\). Here, \(AC\) is opposite angle \(\theta\), so \(AC^{2}=AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot\cos\theta\).

Step2: Substitute the values

Substitute \(AB = 4\), \(BC = 4\), and \(AC = 4\sqrt{3}\) into the formula:

$$ (4\sqrt{3})^{2}=4^{2}+4^{2}-2\times4\times4\times\cos\theta $$

Simplify the left - hand side: \((4\sqrt{3})^{2}=16\times3 = 48\)
Simplify the right - hand side: \(4^{2}+4^{2}-32\cos\theta=16 + 16-32\cos\theta=32-32\cos\theta\)

Step3: Solve for \(\cos\theta\)

We have the equation \(48 = 32-32\cos\theta\)
Subtract 32 from both sides: \(48 - 32=-32\cos\theta\)
\(16=-32\cos\theta\)
Divide both sides by \(- 32\): \(\cos\theta=\frac{16}{-32}=-\frac{1}{2}\)

Step4: Find the angle \(\theta\)

We know that \(\cos\theta=-\frac{1}{2}\) and \(\theta\) is an angle in a triangle (so \(0^{\circ}<\theta<180^{\circ}\)). The angle whose cosine is \(-\frac{1}{2}\) in the interval \((0^{\circ},180^{\circ})\) is \(120^{\circ}\) because \(\cos120^{\circ}=\cos(180^{\circ} - 60^{\circ})=-\cos60^{\circ}=-\frac{1}{2}\)

Answer:

\(120\)