Question

l is in the interior of ∠jkm. find each of the following.

1. m∠jkm if m∠jkl = 42° and m∠lkm = 28°
2. m∠lcm if m∠jkl = 50.4° and m∠jkm = 82.5°

multi - step bd bisects ∠abc. find each of the following.

1. m∠abd if m∠abd=(6x + 4)° and m∠dbc=(8x - 4)°
2. m∠abc if m∠abd=(5y - 3)° and m∠dbc=(3y + 15)°

Explanation:

Step1: Recall angle - addition postulate

The measure of an angle formed by two non - collinear rays with a common endpoint (in this case, if a point lies in the interior of an angle, the measure of the larger angle is the sum of the measures of the two smaller angles).

Step2: Solve for m∠JKM in problem 7

If L is in the interior of ∠JKM, then m∠JKM=m∠JKL + m∠LKM. Given m∠JKL = 42° and m∠LKM=28°, so m∠JKM=42°+28° = 70°.

Step3: Solve for m∠LKM in problem 8

If L is in the interior of ∠JKM, then m∠LKM=m∠JKM - m∠JKL. Given m∠JKL = 50.4° and m∠JKM = 82.5°, so m∠LKM=82.5° - 50.4°=32.1°.

Step4: Recall angle - bisector property

If BD bisects ∠ABC, then m∠ABD=m∠DBC and m∠ABC = 2m∠ABD=2m∠DBC.

Step5: Solve for x in problem 9

Since BD bisects ∠ABC and m∠ABD=(6x + 4)° and m∠DBC=(8x - 4)°, then 6x + 4=8x - 4.
Subtract 6x from both sides: 4 = 2x-4.
Add 4 to both sides: 8 = 2x.
Divide both sides by 2: x = 4.
Then m∠ABD=6x + 4=6×4+4=24 + 4=28°.

Step6: Solve for y in problem 10

Since BD bisects ∠ABC and m∠ABD=(5y - 3)° and m∠DBC=(3y + 15)°, then 5y-3=3y + 15.
Subtract 3y from both sides: 2y-3=15.
Add 3 to both sides: 2y=18.
Divide both sides by 2: y = 9.
Then m∠ABC=2m∠ABD=2(5y - 3)=2(5×9 - 3)=2(45 - 3)=2×42 = 84°.

Answer:

1. 70°
2. 32.1°
3. 28°
4. 84°