Question

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find the smallest angle of $\triangle ijk$.

Explanation:

Step1: Identify triangle parts

We have $\triangle IJK$ with $JK = 64$ in, $IK = 19$ in, and $\angle K = 95^\circ$. We can use the Law of Sines: $\frac{\sin I}{JK}=\frac{\sin J}{IK}=\frac{\sin K}{IJ}$.

Step2: Apply Law of Sines for $\angle J$

First, find $\sin K=\sin 95^\circ\approx0.9962$. Then, $\frac{\sin J}{19}=\frac{\sin 95^\circ}{64}$. So $\sin J=\frac{19\times\sin 95^\circ}{64}\approx\frac{19\times0.9962}{64}\approx\frac{18.9278}{64}\approx0.2958$.

Step3: Find $\angle J$

$\angle J\approx\arcsin(0.2958)\approx17.2^\circ$. Then, $\angle I = 180^\circ - 95^\circ - 17.2^\circ = 67.8^\circ$.

Step4: Compare angles

Angles are $95^\circ$, $67.8^\circ$, $17.2^\circ$. The smallest is $\angle J\approx17^\circ$ (rounded).

Answer:

The smallest angle of $\triangle IJK$ is approximately $17^\circ$ (or more precisely around $17.2^\circ$), so the smallest angle is $\angle J$ with measure approximately $17^\circ$ (using Law of Sines calculations).