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Find $\angle BPC$ below. [asy] pair A = (0,0); pair B = (1,0); pair P = (1/2, sqrt(3)/2); pair D = (0,-1); pair C = (1,-1); draw(A--B--C--D--cycle); draw(A--P--B); draw(rightanglemark(B,A,D, 2)); draw(rightanglemark(C,B,A, 2)); draw(rightanglemark(D,C,B, 2)); draw(rightanglemark(A,D,C, 2)); add(pathticks(P--A, 1, .5, 1, 2)); add(pathticks(P--B, 1, .5, 1, 2)); add(pathticks(A--B, 1, .5, 1, 2)); add(pathticks(B--C, 1, .5, 1, 2)); add(pathticks(C--D, 1, .5, 1, 2)); add(pathticks(A--D, 1, .5, 1, 2)); label("$P$", P, N); label("$A$", A, W); label("$B$", B, E); label("$C$", C, SE); label("$D$", D, SW); [/asy]

 Sep 21, 2019
 #1
avatar+196 
+3

Triangle BPC is an isossilies triangle because sides BP and BC are the same length, so angles BPC and BCP are congruent. To find angle BPC you first need to find angle PBC.

Triangle APB is equilateral so all the angles measure 60 degrees, so angle APB measures 60 degrees. All the angles in a square measure 90 degrees. The measure of angle PBC is the sum of the angles ABC=90 and ABP=60 so PBC=150 degrees.

\(2 \cdot m\angle BPC+m\angle PBC=180\)

150 can be substituted for angle PBC

\(2 \cdot m\angle BPC+150=180\)

subtract 150

\(2 \cdot m\angle BPC=30\)

divide by 2

\(\boxed{m\angle BPC=15}\)

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 Sep 21, 2019
 #3
avatar+2361 
+4

Good job Power27!
 

How were able to decipher the picture code?

CalculatorUser  Sep 22, 2019
 #4
avatar+196 
+3

http://artofproblemsolving.com/texer/

AoPS's TeXeR supports bbCode, LaTeX and Asymptote (what was used to create the drawing).

Asymptote is pretty cool, I've been able to make some fun stuff with it.

power27  Sep 22, 2019

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