Question

∆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 = __

Answer 1

The measures of angles in triangle ABC are found using the distance formula and law of cosines: m∠A ≈ 63.3°, m∠B ≈ 57.1°, and m∠C ≈ 59.6°.

Step-by-step explanation:

To find the measure of each angle in triangle ΔABC with vertices A(-1, 6), B(2, 10), and C (7,-2), we need to first calculate the lengths of the sides of the triangle using the distance formula, and then use the law of cosines to find the angles. The distance formula for two points (x1, y1) and (x2, y2) is √[(x2-x1)^2 + (y2-y1)^2].

Using this formula, we get:

• AB = √[(2 - (-1))^2 + (10 - 6)^2] = √[9 + 16] = √25 = 5
• BC = √[(7 - 2)^2 + (-2 - 10)^2] = √[25 + 144] = √169 = 13
• CA = √[(7 - (-1))^2 + (-2 - 6)^2] = √[64 + 64] = √128 ≈ 11.3

Now, using the law of cosines c^2 = a^2 + b^2 - 2ab*cos(C), we can solve for the angles. For angle A, we have:

cos(A) = (b^2 + c^2 - a^2) / (2bc)

cos(A) = (13^2 + 11.3^2 - 5^2) / (2*13*11.3)

cos(A) ≈ 0.449

A ≈ cos^(-1)(0.449) ≈ 63.3°

Repeating this process for angles B and C:

cos(B) ≈ (5^2 + 11.3^2 - 13^2) / (2*5*11.3)

B ≈ cos^(-1)(cos(B)) ≈ 57.1°

cos(C) ≈ (5^2 + 13^2 - 11.3^2) / (2*5*13)

C ≈ cos^(-1)(cos(C)) ≈ 59.6°

m∠A ≈ 63.3°, m∠B ≈ 57.1°, and m∠C ≈ 59.6°, rounded to one decimal place.