Question

ABC is an equilateral point p is on
base BC such that PC=1/3
BC. If AB = 12 cm
tind AP​

Answer 1

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

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

• → $AP^{2}=40-12$

• → $AP^{2}=28$

• → $AP=2\sqrt{7}cm$

Answer 2

Step-by-step explanation:

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

3

1

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)}

2(AC)(PC)

AC

2

+PC

2

−AP

2

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

2(6)(2)

6

2

+2

2

−AP

2

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

2= 24 40−AP 2

→ AP^{2}=40-12AP 2 =40−12

→ AP^{2}=28AP 2=28

→ AP=2\sqrt{7}cmAP=2√7 cm