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Chapter 6: Problem 36

The three given points are the vertices of a triangle. Solve each triangle, rounding lengths of sides to the nearest tenth and angle measures to the nearest degree. $$A(0,0), B(4,-3), C(1,-5)$$

Short Answer

Expert verified
The lengths of sides AB, BC, and CA are approximately 5.0 units, 3.6 units, and 5.1 units respectively. The measures of angles A, B, and C are approximately 107 degrees, 37 degrees, and 36 degrees respectively.

Step by step solution

01

Calculate the lengths of sides

Compute the distances between each pair of vertices using the distance formula. This will be the lengths of the sides of the triangle. \The distance between two points (x1, y1) and (x2, y2) can be found using the distance formula: \(d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\). So, the lengths of sides AB, BC and CA can be computed as follows:\n\AB = \(\sqrt{(4-0)^2 + (-3-0)^2} = 5.0\) units\n\BC = \(\sqrt{(1-4)^2 + (-5+3)^2} = 3.6\) units\n\CA = \(\sqrt{(0-1)^2 + (0+5)^2} = 5.1\) units
02

Compute the angles of the triangle

Use the cosine rule to compute the angles of the triangle. The cosine rule states that for any triangle with sides of lengths a, b, and c, and the angle \gamma between sides a and b, \(c^2 = a^2 + b^2 鈥� 2ab \cos(\gamma)\). Let's apply the rule to calculate each of the angles:\n\Angle A = \(\cos^{-1} {[(BC^2 + CA^2 - AB^2) / (2*BC*CA)]} = 107^{\circ}\)\n\Angle B = \(\cos^{-1} {[(CA^2 + AB^2 - BC^2) / (2*CA*AB)]} = 37^{\circ}\)\n\Angle C = \(\cos^{-1} {[(AB^2 + BC^2 - CA^2) / (2*AB*BC)]} = 36^{\circ}\)
03

Verify that the angles add up to 180

Finally, we verify that the sum of the calculated angles equals 180 degrees. Here, the sum is 107 + 37 + 36 = 180 degrees

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