Problem:
Distinct points A and B are on a semicircle with diameter MN and center C. The point P is on CN and ∠CAP=∠CBP=10∘. If MA=40∘, then BN equals
Answer Choices:
A. 10∘
B. 15∘
C. 20∘
D. 25∘
E. 30∘
Solution:
In △ACP and △BCP we have (in the order given) the condition a.s.s. Since these triangles are not congruent ( ∠CPA=∠CPB ), we must have that △CPA and ∠CPB are supplementary. From △ACP we compute
∠CPA=180∘−10∘−(180∘−40∘)=30∘.
Thus △CPB=150∘ and BN=△PCB=180∘−10∘−150∘=20∘.
Alternate solution. Again ∠CPA=30∘. Applying the Law of Sines to △ACP and then △BCP, we have
CPsin10∘=ACsin30∘ and CPsin10∘=BCsin∠CPB.
Thus sin∠CPB=21 since AC=BC. As ∠CPB=∠CPA, we must have ∠CPB=150∘. Hence BN=20∘.
