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∆ABC has vertices A(-1, 6), B(2, 10), and C (7,-2). Find the measure of each angle of the triangle. Round decimal answers to the nearest tenth.

M∠A ≈ __
M∠B ≈ __
M∠C ≈ __

1 Answer

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The measures of the angles of the triangle are A = 55.6 degrees, B = 59.5 degrees, and C = 64.9 degrees.

To find the measure of each angle of the triangle, we can use the law of cosines. The law of cosines states that for any triangle ABC,

cos(C) = (a^2 + b^2 - c^2) / (2ab)

where a, b, and c are the side lengths of the triangle opposite angles A, B, and C, respectively.

We can use the distance formula to find the side lengths of the triangle. The distance formula states that the distance between two points (x1, y1) and (x2, y2) is

sqrt((x2 - x1)^2 + (y2 - y1)^2)

Using the distance formula, we find that the side lengths of the triangle are a = 9, b = 13, and c = 10. Substituting these values into the law of cosines, we get

cos(C) = (9^2 + 13^2 - 10^2) / (2 * 9 * 13) = 0.408

Taking the arccosine of both sides, we get

C = arccos(0.408) = 64.9 degrees

We can find the other two angles using the fact that the sum of the angles in a triangle is 180 degrees. Therefore,

A = 180 degrees - B - C = 180 degrees - 59.5 degrees - 64.9 degrees = 55.6 degrees

Therefore, the measures of the angles of the triangle are A = 55.6 degrees, B = 59.5 degrees, and C = 64.9 degrees.

User Neha Choudhary
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