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# Edexcel C2-January 2008,question 5 watch

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1. In triangle ABC , AB=6cm, BC= 13cm and CA=9cm

a) find the value of cos BAC as a fraction in its lowest terms.

Calculations:

a^2=b^2+c^2- 2bc Cos A

13^2=9^2+6^2- 2(9)(6)cosA

169=81+36 - 108 cos A

Cos A = -52/108 = -13/27

b) Show that the area of a triangle ABC is 4√35cm2

Please can anybody show me the whole procedure how to do it?

I know i`ve got to use a formula for area of triangle which is

Area=1/2 bc sinA

But I have no idea how to get the result and whole calculation in a surd form.

Thanks. x
2. okey well you know that cosA = 52/108

so draw a right angled triangle, and fill in the sides (you know that a/h = 52/108, so a = 52 and h = 108). from this you can find the opposite side: 16 root 35.

then you can see that sinA = o/h = (16 root 35) / 108

and simply put this back into the A = 0.5 b c sinA = 0.5 x 9 x 6 x (16 root 35)/108 = 4 root 35.

Edit: just noticed i didn't use the cancelled down fraction, don't know why but yeah it works the same anyway.
3. (Original post by BrightGirl)
okey well you know that cosA = 52/108

so draw a right angled triangle, and fill in the sides (you know that a/h = 52/108, so a = 52 and h = 108). from this you can find the opposite side: 16 root 35.

then you can see that sinA = o/h = (16 root 35) / 108

and simply put this back into the A = 0.5 b c sinA = 0.5 x 9 x 6 x (16 root 35)/108 = 4 root 35.

Edit: just noticed i didn't use the cancelled down fraction, don't know why but yeah it works the same anyway.

Thank you so much, you have been so helpful.
4. (Original post by Janka3112)
Thank you so much, you have been so helpful.
s'ok

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