A side of the triangle below has been extended to form an exterior angle of 132°. Find the value of x...
We know that the sum of the measures of the exterior angles of a triangle is always 360 degrees.
Let's call the angle at point A (the top of the triangle) angle A, the angle at point B angle B, and the angle at point C angle C. We're given that the exterior angle at point A is 132 degrees, so angle A + 132 = 180. Solving for angle A, we find that angle A = 48 degrees.
Now, triangle ABC is an isosceles triangle because two of its sides (AB and AC) are equal in length. This means that angles B and C are also equal. Since the sum of the angles in a triangle is 180 degrees, we can write the equation B + C + 48 = 180.
However, since angles B and C are equal, we can write the equation 2B + 48 = 180. Solving for B, we find that B = 66 degrees.
Since angle A is opposite side BC and angle B is opposite side AC, we can use the Law of Sines to find the value of x. The Law of Sines states that a/sinA = b/sinB = c/sinC, where a, b, and c are the lengths of the sides of a triangle and A, B, and C are the opposite angles.
In this case, we want to find the length of side BC, which is opposite angle A. We can write the equation x/sin48 = AC/sin66. Solving for x, we find that x = AC * sin48 / sin66.