Question

find the measure of ∠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: $a^{2}=b^{2}+c^{2}-2bc\cos(a)$ 34° 41° 51° 56°

Explanation:

Step1: Identify the smallest - angle rule

In a triangle, the smallest angle is opposite the shortest side. Here, the shortest side is 4, so we will find the angle opposite it (let's say $\angle Q$) using the law of cosines. The law of cosines formula is $a^{2}=b^{2}+c^{2}-2bc\cos(A)$. Let $a = 4$, $b = 5$, and $c = 6$.

Step2: Rearrange the law - of - cosines formula

We can rewrite the law of cosines formula for $\cos(A)$ as $\cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}$. Substituting $a = 4$, $b = 5$, and $c = 6$ into the formula, we get $\cos(Q)=\frac{5^{2}+6^{2}-4^{2}}{2\times5\times6}$.

Step3: Calculate the value of $\cos(Q)$

First, calculate the numerator: $5^{2}+6^{2}-4^{2}=25 + 36-16=45$. Then, calculate the denominator: $2\times5\times6 = 60$. So, $\cos(Q)=\frac{45}{60}=0.75$.

Step4: Find the measure of $\angle Q$

To find the measure of $\angle Q$, we use the inverse - cosine function, $\angle Q=\cos^{-1}(0.75)$. Using a calculator, $\angle Q\approx41^{\circ}$.

Answer:

$41^{\circ}$