Let ABC be an isosceles triangle with AB = AC and ∠BAC=30°. Side AB and the median AD contain points Q and P respectively (points P and Q are not at point A) in such a way that PC=PQ. Find m∠PQC.

15°

brainlest

Step-by-step explanation:

Since P is on the median of ΔABC, it is equidistant from points B and C as well as from C and Q. Thus, points B, C, and Q all lie on a circle centered at P. (See the attached diagram.)

The base angles (B and C) of triangle ABC are (180° -30°)/2 = 75°. This means arc QC of the circle centered at P has measure 150°. The diameter of circle P that includes point Q is defined to intersect circle P at R.

Central angle RPC is the difference between arcs QR and QC, so is 180° -150° = 30°. Inscribed angle RQC has half that measure, so is 15°.  Angle PQC has the same measure as angle RQC, so is 15°.

Angle PQC is 15°.

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