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l is in the interior of ∠jkm. find each of the following 7. if m∠jkl = …

Question

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

  1. if m∠jkl = 42° and m∠lkm = 28°, then m∠jkm = _?
  2. if m∠jkl = 56.4° and m∠jkm = 82.5°, then m∠lkm = _?
  3. m∠uow = 50°, and ov bisects ∠uow. what is the m∠voy?

bd bisects ∠abc. find each of the following.

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

Explanation:

Step1: Use angle - addition postulate for question 7

Since $\angle{JKM}=\angle{JKL}+\angle{LKM}$, and $m\angle{JKL} = 42^{\circ}$, $m\angle{LKM}=28^{\circ}$, then $m\angle{JKM}=42^{\circ}+ 28^{\circ}$.
$m\angle{JKM}=70^{\circ}$

Step2: Use angle - addition postulate for question 8

Since $\angle{JKM}=\angle{JKL}+\angle{LKM}$, then $\angle{LKM}=\angle{JKM}-\angle{JKL}$. Given $m\angle{JKL}=56.4^{\circ}$ and $m\angle{JKM}=82.5^{\circ}$, so $m\angle{LKM}=82.5^{\circ}-56.4^{\circ}$.
$m\angle{LKM}=26.1^{\circ}$

Step3: Use angle - bisector property for question 9

If $\overline{OV}$ bisects $\angle{UOW}$, then $\angle{UOV}=\angle{VOW}$. Given $m\angle{UOW} = 50^{\circ}$, so $m\angle{UOV}=\frac{1}{2}m\angle{UOW}=\frac{1}{2}\times50^{\circ}=25^{\circ}$. But it seems there is a mis - label in the question as there is no $\angle{VOY}$ defined clearly. Assuming you want $m\angle{VOW}$, it is $25^{\circ}$

Step4: Use angle - bisector property for question 10

Since $\overline{BD}$ bisects $\angle{ABC}$, then $\angle{ABD}=\angle{DBC}$. So $6x + 4=8x-4$. Solve for $x$:
$8x-6x=4 + 4$, $2x=8$, $x = 4$. Then $m\angle{ABD}=(6\times4 + 4)^{\circ}=28^{\circ}$, $m\angle{DBC}=(8\times4-4)^{\circ}=28^{\circ}$, and $m\angle{ABC}=m\angle{ABD}+m\angle{DBC}=28^{\circ}+28^{\circ}=56^{\circ}$

Step5: Use angle - bisector property for question 11

Since $\overline{BD}$ bisects $\angle{ABC}$, then $\angle{ABD}=\angle{DBC}$. So $5y-3=3y + 15$. Solve for $y$:
$5y-3y=15 + 3$, $2y=18$, $y = 9$. Then $m\angle{ABD}=(5\times9-3)^{\circ}=42^{\circ}$, $m\angle{DBC}=(3\times9 + 15)^{\circ}=42^{\circ}$, and $m\angle{ABC}=m\angle{ABD}+m\angle{DBC}=42^{\circ}+42^{\circ}=84^{\circ}$

Answer:

  1. $70^{\circ}$
  2. $26.1^{\circ}$
  3. $25^{\circ}$
  4. $56^{\circ}$
  5. $84^{\circ}$