MHB Area of an equilateral triangle

[Peking Man]

Given an equilateral triangle ABC, P is any point inside it where PA = 3, PB = 4 and PC = 5.
Find area of the triangle using the Law of Sines or Law of Cosines.

 

[Ackbach]

What ideas have you had so far?

 

[The Chaz]

Given an equilateral triangle ABC, P is any point inside it where PA = 3, PB = 4 and PC = 5.
Find area of the triangle using the Law of Sines or Law of Cosines.

I know you're new around here ;) , so you might not know that it is preferable to show work. (That might be in the rules...)

So I'll get the ball rolling.

Letting a , b, c be the (equal) sides of the "big" triangle and A, B, C the angles formed by... oh dear, I have confused the letters. If the verteces of the outer triangle were renamed...
we have
c^2 = 3^2 + 4^2 - 2(3)(4)cos(C)
etc.
but c^2 = b^2 = a^2 and A + B + C = 360

Can you take it from here?

 

[Prove It]

Heron's Formula may come in handy here, once you have used the Sine Rule and/or Cosine Rule to evaluate the side lengths of the triangle. If a, b, c are the side lengths of your triangle, then

$$ Area = \sqrt{s(s - a)(s - b)(s - c)} $$

where $$ s = \frac{a + b + c}{2} $$

 

[Peking Man]

Heron's Formula may come in handy here, once you have used the Sine Rule and/or Cosine Rule to evaluate the side lengths of the triangle. If a, b, c are the side lengths of your triangle, then

$$ Area = \sqrt{s(s - a)(s - b)(s - c)} $$

where $$ s = \frac{a + b + c}{2} $$

- the teacher said, the Law of Cosines or Law of Sines is enough to solve the problem, but how?

---------- Post added at 11:29 PM ---------- Previous post was at 11:21 PM ----------

Area = (ab/2)sin C = (ac/2)sin B = (bc/2)sin A, and a = b = c. the only missing value is the side of the equilateral triangle ... the application of the Law of Sines and Cosines eludes me so far.

I followed CHAZ suuggestions, but three more internal angles remained unknown ... i have more problems to deal then.

 

[Prove It]

Can you evaluate ONE side of the equilateral triangle? If you have one side, you have them all. Then you can use Heron's Formula (much easier)...

 

[Krizalid1]

This is really a classic problem.
Here's a Hint:

Construct an equilateral triangle on side AP, look for a congruent triangle and use Law of Cosines.