STUDENTGCSE2017
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Consider a triangle ANC such that <BAC is equal to 30 degrees, AB=x, BC=4 and AC=5

Show with your reasoning why x^2 - 5sqrt(3x) +9 =0.

I first drew out the triangle and used the sine rule to get angle B (38.7).
I then subtracted my two angles from 180 degrees to get the final angle and I then did the sine rule once again to find side x (7.45).

I then put 7.45 in the formula above but it did not equal 0 but 41.8... not sure if I’m working this one out wrong. Help please.
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thekidwhogames
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Consider the cosine rule and work backwards to find x.
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STUDENTGCSE2017
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(Original post by thekidwhogames)
Consider the cosine rule and work backwards to find x.
Tried cosine rule still not getting anywhere. Please help me
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thekidwhogames
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(Original post by STUDENTGCSE2017)
Tried cosine rule still not getting anywhere. Please help me
Draw out the triangle and apply cosine rule. That's literally it here.
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STUDENTGCSE2017
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(Original post by thekidwhogames)
Draw out the triangle and apply cosine rule. That's literally it here.
I’m guessing my sides are wrong then? Once I have correctly applied cosine rule, if I submit x into the equation, should it equal 0. Thanks btw
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thekidwhogames
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(Original post by STUDENTGCSE2017)
I’m guessing my sides are wrong then? Once I have correctly applied cosine rule, if I submit x into the equation, should it equal 0. Thanks btw
If you draw out the sides and angle then apply cosine rule. You know the final side, a, and you have the angle, b and c.

So we know a^2 = b^2 +c^2 - 2bcCosA

If you rearrange with this you will get the equation derived.
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Teddyguy
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Sorry I am gonna need a diagram to figure this one out (the way I work)

However I'll give you some tips and tricks.

When you have a triangle with a right angle: use SOH CAH TOA
(Silly Old Hitler Couldn't Advance His Troops Over Africa)

Sine rule: sine A/a = sine B/b = sine C/c
when you want to find a side have small letters up top
when you want an angle have the Sine on top
Use sine when you have an angle and the opposite length

Cosine rule: a^2= b^2+c^2- 2bc cos(A)
use it when you have an angle surrounded by 2 sides or when you have all 3 sides

you can rearrange this to find the angle cos(A)= b^2+c^2-a^2/2bc

learn these rules and you'll never get stuck on a triangle question again
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STUDENTGCSE2017
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(Original post by thekidwhogames)
If you draw out the sides and angle then apply cosine rule. You know the final side, a, and you have the angle, b and c.

So we know a^2 = b^2 +c^2 - 2bcCosA

If you rearrange with this you will get the equation derived.
(Original post by thekidwhogames)
If you draw out the sides and angle then apply cosine rule. You know the final side, a, and you have the angle, b and c.

So we know a^2 = b^2 +c^2 - 2bcCosA

If you rearrange with this you will get the equation derived.
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thekidwhogames
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(Original post by STUDENTGCSE2017)
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Draw your triangle properly and then use the cosine rule as suggested.

b) Use discrimnant and what x represents.
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RDKGames
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(Original post by STUDENTGCSE2017)
Show with your reasoning why x^2 - 5sqrt(3x) +9 =0.
Just so you know, since nobody pointed it out, the middle term should be -5 x \sqrt{3} instead. The x is NOT under the root sign.
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STUDENTGCSE2017
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(Original post by RDKGames)
Just so you know, since nobody pointed it out, the middle term should be -5 x \sqrt{3} instead. The x is NOT under the root sign.
THANKYOU. That is why!
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