Question

In the diagram, $BP$ and $BQ$ trisect $\angle ABC$. $BM$ bisects $\angle PBQ$. Find the ratio of the measure of $\angle MBQ$ to the measure of $\angle ABQ$.

Answer 1

Answer:

1:4

Step-by-step explanation:

If BP and BQ trisect angle ABC, then

$m\angle ABP=m\angle PBQ=m\angle QBC$

If BM bisects the angle MBQ, then

$m\angle PBM=m\angle MBQ=x^{\circ}$

Now, find the measure of angle ABQ in terms of x. First, note that

$m\angle PBQ=2m\angle MBQ=2x^{\circ}$

Now,

$m\angle ABQ=m\angle ABP+m\angle PBQ=2m\angle PBQ=2\cdot 2x^{\circ}=4x^{\circ}$

So, the ratio of the measure of angle MBQ to the measure of angle ABQ is

$\dfrac{x^{\circ}}{4x^{\circ}}=\dfrac{1}{4}$

Answer 2

Answer:

1:4

Step-by-step explanation:

$Let \angle MBQ = x , so \angle MBP=x as well. Therefore, we have \angle PBQ = 2x , so \angle ABP = \angle PBQ = \angle QBC = 2x . Finally, we have \angle ABQ = \angle ABP + \angle PBQ = 4x , so\[\frac{\angle MBQ}{\angle ABQ} = \frac{x}{4x} = \boxed{\frac14}.\]$

Hope this helped! :)