Problem:
Distinct points A and B are on a semicircle with diameter MN and center C. The point P is on CN and β CAP=β CBP=10β. If MA=40β, then BN equals
Answer Choices:
A. 10β
B. 15β
C. 20β
D. 25β
E. 30β
Solution:
In β³ACP and β³BCP we have (in the order given) the condition a.s.s. Since these triangles are not congruent ( β CPAξ =β CPB ), we must have that β³CPA and β CPB are supplementary. From β³ACP we compute
β CPA=180ββ10ββ(180ββ40β)=30β.
Thus β³CPB=150β and BN=β³PCB=180ββ10ββ150β=20β.
Alternate solution. Again β CPA=30β. Applying the Law of Sines to β³ACP and then β³BCP, we have
CPsin10ββ=ACsin30ββ and CPsin10ββ=BCsinβ CPBβ.
Thus sinβ CPB=21β since AC=BC. As β CPBξ =β CPA, we must have β CPB=150β. Hence BN=20β.
