Question

fall 2025 geometry b wwva
right triangle relationships and trigonometry
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
law of cosines: \\( a ^ { 2 } = b ^ { 2 } + c ^ { 2 } - 2 b c \cos ( a ) \\)

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\). In a triangle, the smallest angle is opposite the shortest side. Here, the side opposite \(\angle Q\) is \(PR = 4\). Let \(a = 4\), \(b = 5\), \(c=6\). Then, \(\cos(Q)=\frac{b^{2}+c^{2}-a^{2}}{2bc}\).

Step2: Substitute the values

Substitute \(a = 4\), \(b = 5\), \(c = 6\) into the formula:

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Step3: Find the angle

Since \(\cos(Q)=\frac{3}{4}\), then \(Q=\cos^{-1}(\frac{3}{4})\). Using a calculator, \(Q\approx41.41^{\circ}\). Rounding to the nearest whole degree, \(Q\approx41^{\circ}\).

Answer:

\(41^{\circ}\)