# Perpendicular Bisector of a triangle

Homework Statement:
##ABC## is a triangle such that ##\angle ABC = 37^{\circ}15'##, ##\angle ACB = 59^{\circ}40'##, ##BC = 8## cm; the perpendicular bisector of ##BC## cuts ##BA##, ##CA## produced at ##P, \ Q##. Find the length of ##PQ##.
Relevant Equations:
##\tan(\alpha) = \frac{opposite \ side}{adjacent \ side}##
Here is my attempt to draw a diagram for this problem: I'm confused about the "the perpendicular bisector of ##BC## cuts ##BA##, ##CA## produced at ##P, \ Q##" part of the problem.
How does perpendicular bisector of ##BC## cut the side ##CA##?

mjc123
Homework Helper
It cuts CA produced. Extend the line CA until it meets the bisector. That point is Q (not where you have put it).

• nmnna and FactChecker
It cuts CA produced. Extend the line CA until it meets the bisector. That point is Q (not where you have put it).
Thank you

It cuts CA produced. Extend the line CA until it meets the bisector. That point is Q (not where you have put it). I changed my diagram.
Now I have the right triangle ##\triangle PQC##, where ##CP = 4##cm (since ##PQ## is a perpendicular bisector), ##\angle QCP = 59^{\circ}40'##, so I can find ##PQ## using the relation $$\tan\angle QCP = \frac{PQ}{CP}$$
I got ##\approx 6.818## which is not the answer given in my textbook. Where did I go wrong?

Last edited:
Lnewqban
Gold Member
View attachment 280723
I changed my diagram.
Now I have the right triangle ##\triangle PQC##, where ##CP = 4##cm (since ##PQ## is a perpendicular bisector), ##\angle QCP = 59^{\circ}40'##, so I can find ##PQ## using the relation $$\tan\angle QCP = \frac{PQ}{CP}$$
I got ##\approx 6.818## which is not the answer given in my textbook. Where did I go wrong?
The description seems confusing to me as well.
Could it be that point P should be located where the perpendicular bisector of BC cuts BA?
If it is close to 3.75 cm, then your last diagram is not correct regarding location of P.

• nmnna
mjc123
• 