Question

suppose that $delta klm$ is isosceles with base $lm$.
suppose also that $m\angle l=(4x + 19)^{\circ}$ and $m\angle m=(5x + 12)^{\circ}$,
find the degree measure of each angle in the triangle.
$m\angle k=\square^{\circ}$
$m\angle l=\square^{\circ}$
$m\angle m=\square^{\circ}$

Explanation:

Step1: Set angles equal (base angles)

Since $\Delta KLM$ is isosceles with base $LM$, $\angle L = \angle M$.
$$4x + 19 = 5x + 12$$

Step2: Solve for $x$

Subtract $4x$ and 12 from both sides.
$$19 - 12 = 5x - 4x \implies x = 7$$

Step3: Calculate $\angle L$ and $\angle M$

Substitute $x=7$ into angle expressions.
$$m\angle L = 4(7) + 19 = 28 + 19 = 47^\circ$$
$$m\angle M = 5(7) + 12 = 35 + 12 = 47^\circ$$

Step4: Calculate $\angle K$

Use triangle angle sum ($180^\circ$).
$$m\angle K = 180 - 47 - 47 = 86^\circ$$

Answer:

$m\angle K = 86^\circ$
$m\angle L = 47^\circ$
$m\angle M = 47^\circ$