The vertices of a triangle are a(7, 5), b(4, 2), and c(9, 2). what is m∠abc?

Solution:

Given, the vertices of a triangle are

a(7, 5), b(4, 2), c(9, 2)

We have to find the measure of angle abc.

Using distance formula to calculate the length of each side of the triangle.

Distance = \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

Length of side ab = \(\sqrt{(4-7)^{2}+(2-5)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}\\ab=\sqrt{9+9}=\sqrt{18}\)

Length of side bc = \(\sqrt{(9-4)^{2}+(2-2)^{2}}=\sqrt{(5)^{2}+(0)^{2}}\\bc=\sqrt{25}=5\)

Length of side ac = \(\sqrt{(9-7)^{2}+(2-5)^{2}}=\sqrt{(2)^{2}+(-3)^{2}}\\bc=\sqrt{4+9}=\sqrt{13}\)

Using cosine law,

\(cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}\)

\(cos B = \frac{(5)^{2}+(\sqrt{18})^{2}-(\sqrt{13})^{2}}{2(5)(\sqrt{18})}\)

\(cos B = \frac{25+18-13}{10(\sqrt{18})}\)

\(cos B = \frac{25+5}{10(\sqrt{18})}\)

\(cos B = \frac{30}{10(\sqrt{18})}\)

\(cos B = \frac{3}{\sqrt{18}}\)

\(cos B = \frac{3}{\sqrt{9\times 2}}=\frac{3}{3\sqrt{2}}\)

\(cos B = \frac{1}{\sqrt{2}}\)

Taking inverse,

B = cos^-1(1/√2)

B = 45°

Therefore, the measure of angle abc is 45°.

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The vertices of a triangle are a(7, 5), b(4, 2), and c(9, 2). what is m∠abc?

Summary:

The vertices of a triangle are a(7, 5), b(4, 2), and c(9, 2). m∠abc is 45°.