Question

∆ABC is an equilateral triangle. Point P is on base BC such that

PC = 1/3BC. If AB = 12 cm, find AP.​

Answer 1

Answer:

Step-by-step explanation:

It is given that ABC is an equilateral triangle, then AB=BC=AC=6cm and ∠A=∠B=∠C=60°, then according to question, PC$=\frac{1}{3}$

therefore PC=2 cm.

Now, using the cosine formula in ΔAPC, we have

$cos∠C= \frac{AC^{2}+PC^{2}-AP^{2}}{2(AC)(PC)}$

cos60°=$\frac{6^{2}+2^{2}-AP^{2}}{2(6)(2)}$

$\frac{1}{2}=\frac{40-AP^{2}}{24}$

$AP^{2}$

$AP^{2}$

$AP=2\sqrt{7}cmAP=2$