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Maths Trig Help!

Can someone tell me how to do these 2 questions? For 6, I tried doing cosine rule to find x but didn’t work! And I don’t even know how to start 7! 5864569C-E6C2-4099-84EA-9EFCDD5EEDFE.jpg.jpeg 505AB131-DBAD-4254-85A5-F4895068ABE0.jpg.jpeg
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by alevels2020
Can someone tell me how to do these 2 questions? For 6, I tried doing cosine rule to find x but didn’t work! And I don’t even know how to start 7! 5864569C-E6C2-4099-84EA-9EFCDD5EEDFE.jpg.jpeg 505AB131-DBAD-4254-85A5-F4895068ABE0.jpg.jpeg


Can you show your working for 6? I would have thought the cosine rule was the obvious thing to do.
Original post by mqb2766
Can you show your working for 6? I would have thought the cosine rule was the obvious thing to do.


Yes, sorry it’s a bit messy I did it a couple of times9949FD5D-D583-46E9-B52E-D31B88878BB4.jpg.jpeg also ignore the top part where it says finding errors... this is just scrap paper aha
Ok, i might be wrong but if you look carefully you can apply the cosine rule to find side a. From there you can apply the sine rule to find angle B and the use the fact that angles on a straight line add up to 180. This should be the right method for question 7
Reply 4
Avatar for Sinnoh
Sinnoh
22
For 6 you use the cosine rule to get a quadratic equation in terms of x. Check your work for mistakes - they can happen, since this question is quite algebra-heavy.

As for 7, in part a you simply treat the whole thing as a triangle and use the cosine rule once again to find the unknown length. For part b you would need to calculate the value of angle ABC which will be slightly easier to do with the sine rule once you've done part a. Remember that the sum of angles on a straight line is 180 degrees.
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by alevels2020
Yes, sorry it’s a bit messy I did it a couple of times9949FD5D-D583-46E9-B52E-D31B88878BB4.jpg.jpeg also ignore the top part where it says finding errors... this is just scrap paper aha

Its hard to read in the image, but is -2bccos(120) right?
cos(120) = -0.5
bc = x^2-1
-2bccos(120) = x^2-1
Original post by sqrt of 5
Ok, i might be wrong but if you look carefully you can apply the cosine rule to find side a. From there you can apply the sine rule to find angle B and the use the fact that angles on a straight line add up to 180. This should be the right method for question 7


Ah ok, thank u!! That makes sense
Original post by Sinnoh
For 6 you use the cosine rule to get a quadratic equation in terms of x. Check your work for mistakes - they can happen, since this question is quite algebra-heavy.

As for 7, in part a you simply treat the whole thing as a triangle and use the cosine rule once again to find the unknown length. For part b you would need to calculate the value of angle ABC which will be slightly easier to do with the sine rule once you've done part a. Remember that the sum of angles on a straight line is 180 degrees.


Thank you!
Original post by mqb2766
Its hard to read in the image, but is -2bccos(120) right?
cos(120) = -0.5
bc = x^2-1
-2bccos(120) = x^2-1


ohh I think I’ve added instead of multiply, thank you!
Reply 9
Avatar for mqb2766
mqb2766
21
Original post by alevels2020
ohh I think I’ve added instead of multiply, thank you!


Yeah, you've got the right approach, just check your working.

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