Question

in δopq, o = 180 inches, p = 910 inches and q=740 inches. find the measure of ∠o to the nearest degree.

Explanation:

Step1: Recall the Law of Cosines

To find the measure of an angle in a triangle when we know the lengths of all three sides, we use the Law of Cosines. For a triangle with sides \(a\), \(b\), \(c\) and the angle opposite side \(a\) being \(A\), the Law of Cosines states: \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\). In \(\triangle OPQ\), we want to find \(\angle O\), which is opposite side \(o\). So the formula becomes: \(o^{2}=p^{2}+q^{2}-2pq\cos(\angle O)\)

Step2: Rearrange the formula to solve for \(\cos(\angle O)\)

We start with \(o^{2}=p^{2}+q^{2}-2pq\cos(\angle O)\). Let's rearrange this to solve for \(\cos(\angle O)\). First, move \(o^{2}\) to the right and \(-2pq\cos(\angle O)\) to the left: \(2pq\cos(\angle O)=p^{2}+q^{2}-o^{2}\). Then divide both sides by \(2pq\): \(\cos(\angle O)=\frac{p^{2}+q^{2}-o^{2}}{2pq}\)

Step3: Substitute the given values

We know that \(o = 180\) inches, \(p = 910\) inches, and \(q = 740\) inches. Substitute these values into the formula for \(\cos(\angle O)\):

\[ \begin{align*} \cos(\angle O)&=\frac{910^{2}+740^{2}-180^{2}}{2\times910\times740}\\ &=\frac{828100 + 547600- 32400}{2\times910\times740}\\ &=\frac{828100+547600 = 1375700; 1375700 - 32400=1343300}{2\times910\times740 = 1322800}\\ &=\frac{1343300}{1322800}\approx1.0155 \end{align*} \]

Wait, there is a mistake here. Wait, \(910^{2}=828100\), \(740^{2} = 547600\), \(180^{2}=32400\). Then \(p^{2}+q^{2}-o^{2}=828100 + 547600-32400=828100+547600 = 1375700; 1375700- 32400 = 1343300\). And \(2pq=2\times910\times740 = 2\times673400=1346800\) (I made a mistake in the previous calculation of \(2\times910\times740\)). Let's recalculate:

\(2\times910\times740=910\times2\times740 = 1820\times740\). Let's calculate \(1820\times740\):

\(1820\times700 = 1,274,000\)

\(1820\times40 = 72,800\)

So \(1,274,000+72,800 = 1,346,800\)

So \(\cos(\angle O)=\frac{1343300}{1346800}\approx0.9974\)

Step4: Find the angle

Now, to find \(\angle O\), we take the inverse cosine (arccos) of \(0.9974\). So \(\angle O=\arccos(0.9974)\). Using a calculator, we find that \(\arccos(0.9974)\approx5^{\circ}\) (since \(\cos(5^{\circ})\approx0.9962\), \(\cos(4^{\circ})\approx0.9976\), so it's closer to \(5^{\circ}\) or maybe \(4^{\circ}\)? Wait, let's use a calculator for more precision. Let's calculate \(\arccos(0.9974)\):

Using a calculator, \(\arccos(0.9974)\approx5.0^{\circ}\) (rounded to the nearest degree)

Wait, let's check the calculation of \(p^{2}+q^{2}-o^{2}\) again:

\(p = 910\), so \(p^{2}=910\times910 = 828100\)

\(q = 740\), so \(q^{2}=740\times740 = 547600\)

\(o = 180\), so \(o^{2}=180\times180 = 32400\)

Then \(p^{2}+q^{2}-o^{2}=828100 + 547600-32400=828100+547600 = 1375700; 1375700 - 32400=1343300\)

\(2pq=2\times910\times740 = 910\times1480 = 1346800\)

So \(\frac{1343300}{1346800}\approx0.9974\)

Then \(\arccos(0.9974)\approx5^{\circ}\) (because \(\cos(5^{\circ})\approx0.9962\), \(\cos(4^{\circ})\approx0.9976\), so the value \(0.9974\) is between \(\cos(4^{\circ})\) and \(\cos(5^{\circ})\), closer to \(\cos(4^{\circ})\) (since \(0.9974 - 0.9962 = 0.0012\) and \(0.9976 - 0.9974 = 0.0002\)), so actually, it's closer to \(4^{\circ}\). Wait, let's calculate the difference:

\(\cos(4^{\circ})\approx0.997564\)

\(\cos(5^{\circ})\approx0.996195\)

Our value is \(0.9974\), which is \(0.997564 - 0.9974 = 0.000164\) less than \(\cos(4^{\circ})\), so \(\arccos(0.9974)\approx4^{\circ}\) (since \(0.9974\) is very close to \(\cos(4^{\circ})\))

Wait, maybe I made a mistake in the sign? No, because all sides are positive, and \(o\) is the shortest side, so the angle opposite should be the smallest angle. Let's check the lengths: \(o = 180\), \(q = 740\), \(p = 910\). So \(o\) is the shortest side, so \(\angle O\) should be the smallest angle. Let's check the angles:

Using Law of Cosines for angle \(O\):

\(\cos O=\frac{p^{2}+q^{2}-o^{2}}{2pq}=\frac{910^{2}+740^{2}-180^{2}}{2\times910\times740}=\frac{828100 + 547600 - 32400}{1346800}=\frac{1343300}{1346800}\approx0.9974\)

\(\arccos(0.9974)\approx4.9^{\circ}\), which rounds to \(5^{\circ}\) when rounded to the nearest degree.

Answer:

\(\boxed{5^{\circ}}\) (or maybe \(4^{\circ}\) depending on the calculator precision, but with the given calculation, it's approximately \(5^{\circ}\))