Question

law of cosines
a surveyor measures the lengths of the sides of a triangular plot of land. what is the measure of the angle of the triangular plot at which the surveyor stands? approximate to the nearest degree.
$cos^{-1}(0.563) = 56^circ$
$cos^{-1}(0.75) = 41^circ$
$cos^{-1}(0.125) = 83^circ$

Explanation:

Step1: Recall Law of Cosines

The Law of Cosines for a triangle with sides \(a\), \(b\), \(c\) and the angle \(\theta\) opposite side \(c\) is \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\). We can rearrange it to solve for \(\cos\theta\): \(\cos\theta=\frac{a^{2}+b^{2}-c^{2}}{2ab}\). Here, the sides adjacent to the angle where the surveyor stands are \(a = 200\) m, \(b=250\) m, and the side opposite is \(c = 300\) m.

Step2: Substitute values into the formula

Substitute \(a = 200\), \(b = 250\), \(c=300\) into \(\cos\theta=\frac{a^{2}+b^{2}-c^{2}}{2ab}\).
First, calculate \(a^{2}=200^{2} = 40000\), \(b^{2}=250^{2}=62500\), \(c^{2}=300^{2}=90000\).
Then, \(a^{2}+b^{2}-c^{2}=40000 + 62500-90000=12500\).
And \(2ab=2\times200\times250 = 100000\).
So \(\cos\theta=\frac{12500}{100000}=0.125\)? Wait, no, wait, maybe I mixed up the sides. Wait, the angle where the surveyor stands: the two sides from the surveyor are 200 and 250, and the opposite side is 300? Wait, no, let's re - check. Wait, maybe the sides are \(a = 200\), \(b = 250\), and the included angle is \(\theta\), with the third side \(c = 300\). So using Law of Cosines: \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\). So \(300^{2}=200^{2}+250^{2}-2\times200\times250\times\cos\theta\).
\(90000=40000 + 62500-100000\cos\theta\)
\(90000=102500-100000\cos\theta\)
\(100000\cos\theta=102500 - 90000\)
\(100000\cos\theta=12500\)
\(\cos\theta=\frac{12500}{100000}=0.125\)? Wait, that gives \(\cos^{- 1}(0.125)\approx83^{\circ}\). But wait, maybe I made a mistake in side assignment. Wait, another way: if the sides are 200, 250, 300, and we want the angle between 200 and 250, then \(c = 300\), \(a = 200\), \(b = 250\). So the formula for \(\cos\theta\) (angle between \(a\) and \(b\)) is \(\cos\theta=\frac{a^{2}+b^{2}-c^{2}}{2ab}\). So \(\cos\theta=\frac{200^{2}+250^{2}-300^{2}}{2\times200\times250}=\frac{40000 + 62500-90000}{100000}=\frac{12500}{100000}=0.125\). Then \(\theta=\cos^{-1}(0.125)\approx83^{\circ}\). Wait, but let's check the other options. Wait, maybe I assigned the sides wrong. Wait, maybe the sides are 200, 300, and 250, and the angle between 200 and 300. Let's try that. Let \(a = 200\), \(b = 300\), \(c = 250\). Then \(\cos\theta=\frac{200^{2}+300^{2}-250^{2}}{2\times200\times300}=\frac{40000+90000 - 62500}{120000}=\frac{67500}{120000}=0.5625\approx0.563\). Then \(\cos^{-1}(0.563)\approx56^{\circ}\). Ah, I see, I assigned the wrong side as \(c\) earlier. The side opposite the angle we want: if the surveyor is at the vertex with sides 200 and 300, and the opposite side is 250. So let's correct that. Let \(a = 200\), \(b = 300\), \(c = 250\) (the side opposite the angle \(\theta\) at the surveyor's position). Then using Law of Cosines: \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\). So \(250^{2}=200^{2}+300^{2}-2\times200\times300\times\cos\theta\). \(62500 = 40000+90000-120000\cos\theta\). \(62500=130000 - 120000\cos\theta\). \(120000\cos\theta=130000 - 62500\). \(120000\cos\theta = 67500\). \(\cos\theta=\frac{67500}{120000}=0.5625\approx0.563\). Then \(\theta=\cos^{-1}(0.563)\approx56^{\circ}\).

Answer:

\(\cos^{-1}(0.563)=56^{\circ}\) (the first option)