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3 side of a triangle; find an angle; not right angle.

Look at Q 24 on this paper: 1st Post, attachment. The ordering isn't chronological, it's 3rd to last.

I can't see a way of doing it, as it's not right angle and you need at least one angle to work out another angle; don't you?
Reply 1
Avatar for gdunne42
gdunne42
21
Original post by Hiya123321
Look at Q 24 on this paper: 1st Post, attachment. The ordering isn't chronological, it's 3rd to last.

I can't see a way of doing it, as it's not right angle and you need at least one angle to work out another angle; don't you?


Cosine rule - on the formula page
Reply 2
Avatar for Hiya123321
Hiya123321
OP
Original post by gdunne42
Cosine rule - on the formula page


But you only get

a^2 = b^2 + c^2 - 2bc * cos A,
not CosA = ...........
Reply 3
Avatar for gdunne42
gdunne42
21
Original post by Hiya123321
But you only get

a^2 = b^2 + c^2 - 2bc * cos A,
not CosA = ...........


correct

and for the sine rule you are only given a/sinA = b/sinB .........

and it's useful to know that Sin/a = Sin B/b ........

while most recent questions seem to have asked you to find a side given 2 sides and an angle, Cos rule can also be used to find an angle given 3 sides.

use your 'make CosA the subject of the formula' skills to rearrange it
(edited 12 years ago)
Reply 4
Avatar for sagskay
sagskay
a^2-b^2-c^2/-2bc = cos A
so cos-1(ANS) = A.
Reply 5
Avatar for nuodai
nuodai
17
Original post by Hiya123321
But you only get

a^2 = b^2 + c^2 - 2bc * cos A,
not CosA = ...........


Indeed, but you can rearrange an equation, right? Plug in the values of a,b,c and then rearrange to get cos⁡A=stuff\cos A = \text{stuff}.
Reply 6
Avatar for Hiya123321
Hiya123321
OP
Thanks.

The OP who made the question paper:

Sub'd the values, so he had:

225 = 144 + 100 - 240 * CosA

Then swapped the -240CosA with 225

240 * CosA = 244 - 225

Then continued solving it:

CosA = 9 / 240

CosA = 0.0375

Then did cos-1

Which makes the answer 87.85

Good luck, if you've got the exam tomorrow.

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