ABC is an equilateral point p is onbase BC such that PC=1/3BC. If AB =

1.	ABC is an equilateral point p is onbase BC such that PC=1/3BC. If AB = 12 cmtind AP
Answer» GIVEN:- ABC is an EQUILATERAL triangle. AB = BC = AC = 6cm. ∠A=∠B=∠C=60°. step-by-step EXPLAINATION:- according to question, → PC= $\frac{1}{3}BC$ therefore PC=2 cm. Now, using the cosine formula in ΔAPC, we have → cos∠C= $\sf\frac{AC^{2}+PC^{2}-AP^{2}}{2(AC)(PC)}$ → cos60°= $\sf\frac{6^{2}+2^{2}-AP^{2}}{2(6)(2)}$ → $\sf\frac{1}{2}=\sf\frac{40-AP^{2}}{24}$ → $AP^{2}=40-12$ → $AP^{2}=28$ → $AP=2\sqrt{7}cm$