Question

Consider a triangle ABC like the one below. Suppose that A=124, b=72. and c=43. (The figure is not drawn to scale.) Solve the triangle Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. If there is more than one solution, use the button labeled "

Answer 1

B = 35.7°

C = 20.3°

a = 102.5

Explanation:

Given that angle A is included between both sides b and c, it's best to use the Law of Cosines to solve the rest of the triangle. First, let's determine the length of side a since that is easiest:

`a=√(b^2+c^2-2bc\cos(A))\\a=√(72^2+43^2-2(72)(43)\cos(124))\\a\approx102.5`

Now that we have all the side lengths, we can solve for either one of the unknown angles and then complete the triangle:

`\cos(C)=(a^2+b^2-c^2)/(2ab)\\\\\cos(C)=(102.5^2+72^2-43^2)/(2(102.5)(72))\\\\C=\cos^(-1)((102.5^2+72^2-43^2)/(2(102.5)(72)))\\\\C\approx20.3^\circ`

Getting the last angle is pretty simple since it's just Triangle Angle-Sum Theorem:

`m\angle A+m\angle B+m\angle C=180^\circ\\124+B+20.3=180\\144.3+B=180\\m\angle B=35.7^\circ`