Question

Right triangle ABC is graphed in the xy coordinate plane with vertices at (0,0) , (5, 0) and (5, 12) If is the angle in the triangle at the origin, what is the cos(pi + theta)?

Answer 1

Explanation:

ince right triangle ABC is graphed in the xy-coordinate plane with vertices at (0,0), (5,0) and (5,12), we can use the coordinates of the vertices to find the lengths of the sides of the triangle.

The length of side AB is 5 units, the length of side BC is 12 units, and the length of side AC is given by the Pythagorean theorem:

AC^2 = AB^2 + BC^2

AC^2 = 5^2 + 12^2

AC^2 = 169

AC = 13

Therefore, right triangle ABC is a 5-12-13 triangle.

The angle at the origin is opposite to the side of length 12, so it is the angle opposite to the side BC. We can find this angle using the inverse tangent function:

tan(theta) = opposite / adjacent

tan(theta) = 12 / 5

theta = tan^-1(12/5) ≈ 68.2°

Since cos(pi + theta) = -cos(theta), we have:

cos(pi + theta) = -cos(theta) = -cos(tan^-1(12/5))

Using the identity cos(arctan(x)) = 1 / sqrt(1 + x^2), we can simplify this expression:

cos(tan^-1(12/5)) = 1 / sqrt(1 + (12/5)^2) = 5 / 13

Therefore, cos(pi + theta) = -cos(theta) = -cos(tan^-1(12/5)) = -5/13.