Question

1. (mangle ij f = 4x + 10), (mangle fjk = 80^{circ}), and (mangle ijk = 12x + 10). find (x). 13) find (x) if (mangle blm = 140^{circ}), (mangle klm = 165x - 1), and (mangle klb = 25x - 1).

Explanation:

Step1: Use angle - addition postulate

$m\angle IJF + m\angle FJK=m\angle IJK$, so $(4x + 10)+80=12x + 10$.

Step2: Simplify the equation

$4x+90 = 12x + 10$.

Step3: Solve for x

$90-10=12x - 4x$, $80 = 8x$, $x = 10$.

Step1: Use angle - addition postulate

$m\angle KLB+m\angle BLM=m\angle KLM$, so $(25x - 1)+140=165x - 1$.

Step2: Simplify the equation

$25x+139 = 165x - 1$.

Step3: Solve for x

$139 + 1=165x - 25x$, $140 = 140x$, $x = 1$.

Answer:

$x = 10$