Question

1. if δjkl is isosceles, what is m∠jkl?

(10x - 13)° at j, (3x + 22)° at l, triangle jkl with k at top. options: a. 37°, b. 74°, c. 106°, d. 143°

Explanation:

Step1: Set angles equal (isosceles)

In isosceles \(\triangle JKL\), \(\angle J = \angle L\) (base angles). So \(10x - 13 = 3x + 22\).

Step2: Solve for \(x\)

Subtract \(3x\) and add \(13\): \(10x - 3x = 22 + 13\) → \(7x = 35\) → \(x = 5\).

Step3: Find \(\angle J\) (or \(\angle L\))

Substitute \(x = 5\) into \(\angle J\): \(10(5) - 13 = 50 - 13 = 37^\circ\). So \(\angle J = \angle L = 37^\circ\).

Step4: Find \(\angle JKL\)

Sum of angles in a triangle: \(180^\circ\). So \(m\angle JKL = 180 - 37 - 37 = 106^\circ\).

Answer:

C. \(106^\circ\)