Question

find the measure of $\angle q$, the smallest angle in a triangle whose sides have lengths 4, 5, and 6. round the measure to the nearest whole degree. \bigcirc $34^\circ$ \bigcirc $41^\circ$ \bigcirc $51^\circ$ \bigcirc $56^\circ$ law of cosines: $a^2 = b^2 + c^2 - 2bc\cos(a)$

Explanation:

Step1: Identify sides for ∠Q

In triangle \(PQR\), for \(\angle Q\), the sides adjacent are \(PQ = 5\), \(QR = 6\), and the side opposite is \(PR = 4\). Using the Law of Cosines \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\), here \(a = 4\), \(b = 5\), \(c = 6\), and \(A=\angle Q\).

Step2: Substitute into Law of Cosines

Substitute into the formula: \(4^{2}=5^{2}+6^{2}-2\times5\times6\times\cos(\angle Q)\)
Simplify: \(16 = 25 + 36-60\cos(\angle Q)\)

Step3: Solve for \(\cos(\angle Q)\)

Rearrange: \(60\cos(\angle Q)=25 + 36 - 16\)
\(60\cos(\angle Q)=45\)
\(\cos(\angle Q)=\frac{45}{60}=0.75\)

Step4: Find \(\angle Q\)

Take the arccosine: \(\angle Q=\arccos(0.75)\approx41^{\circ}\)

Answer:

\(41^{\circ}\) (corresponding to the option: 41°)