Question

find the measure of ∠j, the smallest angle in a triangle with sides measuring 11, 13, and 19. round to the nearest whole degree. 30° 34° 42° 47° law of cosines: a² = b² + c² - 2bc cos(a)

Explanation:

Step1: Recall the law of cosines

The smallest angle is opposite the smallest side. So, for $\angle J$ opposite side $KL = 11$, using the law of cosines $a^{2}=b^{2}+c^{2}-2bc\cos(A)$. Here $a = 11$, $b = 13$, $c = 19$. Then $11^{2}=13^{2}+19^{2}-2\times13\times19\times\cos(J)$.

Step2: Simplify the equation

$121=169 + 361-494\cos(J)$.
$121=530-494\cos(J)$.
$494\cos(J)=530 - 121$.
$494\cos(J)=409$.

Step3: Solve for $\cos(J)$

$\cos(J)=\frac{409}{494}\approx0.828$.

Step4: Find the angle

$J=\cos^{-1}(0.828)$. Using a calculator, $J\approx34^{\circ}$.

Answer:

$34^{\circ}$