Question

find the measure of $angle j$, the smallest angle in a triangle with sides measuring 11, 13, and 19. round to the nearest whole degree.
$\bigcirc$ $30^circ$
$\bigcirc$ $34^circ$
$\bigcirc$ $42^circ$
$\bigcirc$ $47^circ$
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 11, opposite $\angle J$. Assign values for the Law of Cosines: $a=11$, $b=13$, $c=19$, $A=\angle J$.

Step2: Rearrange Law of Cosines

Isolate $\cos(A)$ to solve for the angle.
$$\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}$$

Step3: Substitute values into formula

$$\cos(\angle J) = \frac{13^2 + 19^2 - 11^2}{2 \times 13 \times 19}$$

Step4: Calculate numerator and denominator

$$13^2=169, 19^2=361, 11^2=121$$
$$\text{Numerator}=169+361-121=409$$
$$\text{Denominator}=2 \times 13 \times 19=494$$
$$\cos(\angle J)=\frac{409}{494} \approx 0.8279$$

Step5: Find inverse cosine

$$\angle J = \cos^{-1}(0.8279) \approx 34^\circ$$

Answer:

34°