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. 34° 41° 51° 56° law of cosines: a² = b² + c² - 2bc cos(a)

Explanation:

Step1: Identify the smallest - side opposite the smallest angle

The smallest side is 4, and the angle opposite it is $\angle Q$. Let $a = 4$, $b = 5$, $c = 6$.

Step2: Apply the law of cosines

The law of cosines formula is $a^{2}=b^{2}+c^{2}-2bc\cos(A)$. Substituting the values, we get $4^{2}=5^{2}+6^{2}-2\times5\times6\times\cos(Q)$.
So, $16 = 25 + 36-60\cos(Q)$.

Step3: Simplify the equation

$16=61 - 60\cos(Q)$.
Then, $60\cos(Q)=61 - 16$.
$60\cos(Q)=45$.

Step4: Solve for $\cos(Q)$

$\cos(Q)=\frac{45}{60}=\frac{3}{4}$.

Step5: Find the measure of $\angle Q$

$Q=\cos^{-1}(\frac{3}{4})\approx41^{\circ}$.

Answer:

$41^{\circ}$