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Question
triangle mnp has vertices n(2,2) and p(0,-2). the triangle is symmetric about the y - axis. what is the approximate measure of the largest angle in the triangle? 52.8° 60.0° 63.6° 82.9°
Step1: Find the coordinates of point \(M\)
Since the triangle is symmetric about the \(y -\)axis, if \(N(2,2)\), then \(M(- 2,2)\) (because for a point \((x,y)\) symmetric about the \(y -\)axis, the symmetric point is \((-x,y)\)). And \(P(0,-2)\)
Step2: Calculate the lengths of the sides using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
- Length of \(MN\): \(d_{MN}=\sqrt{(2 + 2)^2+(2 - 2)^2}=\sqrt{16}=4\)
- Length of \(MP\): \(d_{MP}=\sqrt{(-2-0)^2+(2 + 2)^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.47\)
- Length of \(NP\): \(d_{NP}=\sqrt{(2-0)^2+(2 + 2)^2}=\sqrt{4 + 16}=\sqrt{20}\approx4.47\)
Step3: Use the Law of Cosines \(c^{2}=a^{2}+b^{2}-2ab\cos C\)
Let \(a = 4\), \(b = 4.47\), \(c = 4.47\). We want to find the angle opposite the longest side (but in this case \(MP = NP\)). Let's find the angle at \(M\).
\(M=\cos^{-1}(0.447)\approx63.6^{\circ}\)
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\(63.6^{\circ}\)