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: (a^2 = b^2 + c^2 - 2bccos(a))

Explanation:

Step1: Match side to angle

In a triangle, the smallest angle is opposite the shortest side. The shortest side is 4, opposite $\angle Q$. Assign:
$a=4$, $b=5$, $c=6$, $A=\angle Q$

Step2: Rearrange Law of Cosines

Isolate $\cos(A)$ from $a^2 = b^2 + c^2 - 2bc\cos(A)$:
$\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}$

Step3: Substitute values

Plug in $a=4$, $b=5$, $c=6$:
$\cos(\angle Q) = \frac{5^2 + 6^2 - 4^2}{2\times5\times6} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = 0.75$

Step4: Calculate angle

Find the inverse cosine of 0.75:
$\angle Q = \cos^{-1}(0.75) \approx 41.41^\circ$

Step5: Round to nearest degree

Round $41.41^\circ$ to the whole number:
$\angle Q \approx 41^\circ$

Answer:

41°