Question

what is the measure of $angle p$ to the nearest whole degree?$\bigcirc$ $35^circLXB0\bigcirc$ $57^circ$$\bigcirc$ $72^circ$law of cosines: $a^2 = b^2 + c^2 - 2bccos(a)$

Explanation:

Step1: Assign values to Law of Cosines

In $\triangle PQR$, for $\angle P$, side opposite is $QR=15$, adjacent sides are $PQ=17$, $PR=14$.
Law of Cosines: $a^2 = b^2 + c^2 - 2bc\cos(A)$ becomes:
$$15^2 = 17^2 + 14^2 - 2(17)(14)\cos(P)$$

Step2: Calculate squared side lengths

Compute each squared term:
$$225 = 289 + 196 - 476\cos(P)$$

Step3: Simplify right-hand side

Sum the constants:
$$225 = 485 - 476\cos(P)$$

Step4: Isolate the cosine term

Rearrange to solve for $\cos(P)$:
$$476\cos(P) = 485 - 225$$
$$476\cos(P) = 260$$
$$\cos(P) = \frac{260}{476} \approx 0.5462$$

Step5: Find $\angle P$ using arccosine

Calculate the inverse cosine:
$$P = \arccos(0.5462) \approx 57^\circ$$

Answer:

57° (Option C)