Question

in δopq, o = 290 inches, p = 720 inches and q=600 inches. find the measure of ∠p to the nearest 10th of a degree.

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(A\) opposite side \(a\) is \(a^{2}=b^{2}+c^{2}-2bc\cos A\). In \(\triangle OPQ\), angle \(P\) is opposite side \(p\), so the formula becomes \(p^{2}=o^{2}+q^{2}-2oq\cos P\). We need to solve for \(\cos P\) first.
Rearranging the formula for \(\cos P\):
\[
\cos P=\frac{o^{2}+q^{2}-p^{2}}{2oq}
\]

Step2: Substitute the given values

Given \(o = 290\) inches, \(p=720\) inches, \(q = 600\) inches. Substitute these values into the formula for \(\cos P\):
\[
\cos P=\frac{290^{2}+600^{2}-720^{2}}{2\times290\times600}
\]
First, calculate the numerator:
\(290^{2}=84100\), \(600^{2}=360000\), \(720^{2}=518400\)
\[

$$\begin{align*} \text{Numerator}&=84100 + 360000-518400\\ &=444100 - 518400\\ &=- 74300 \end{align*}$$

\]
Denominator: \(2\times290\times600 = 348000\)
So, \(\cos P=\frac{-74300}{348000}\approx - 0.2135\)

Step3: Find the angle \(P\)

To find \(P\), we take the inverse cosine (arccos) of \(- 0.2135\):
\[
P=\arccos(-0.2135)
\]
Using a calculator, make sure it is in degree mode. \(\arccos(-0.2135)\approx102.3^{\circ}\) (rounded to the nearest tenth of a degree)

Answer:

\(102.3^{\circ}\)