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°

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°.

(x) - (y) = (1 + 4i)

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube