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#### Peking Man

##### New member

- Jan 27, 2012

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Find area of the triangle using the Law of Sines or Law of Cosines.

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- Jan 27, 2012

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Find area of the triangle using the Law of Sines or Law of Cosines.

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What ideas have you had so far?

- Jan 26, 2012

- 24

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...)

Find area of the triangle using the Law of Sines or Law of Cosines.

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?

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

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

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- Jan 27, 2012

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

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

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

---------- 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.

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Here's a Hint:

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