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AS Level Maths Edexcel Trig Help URGENT

Alright so I have this question and I want to see how others have answered it.

The diagram shows triangle ABC with AC=8x-3, BC=4x-1, angle ABC=120 degrees and angle ACB=15 degrees.

(a) Show that the exact value of x is 9+√6/20 (7 marks)
(b) Find the area of triangle ABC giving your answer to 2 decimal places
(3 marks)
I don't mind anyone answering it please. This question has been bugging me for so long
What have you been able to do so far?
Not much other than find the other remaining angle. I'm unsure as to how I should start it
I'm seeing the first part u use the sine rule or cosine rule and then u simplify to make ur answer look like What They want then second one u just use replace the value of x in the equation for the length is AB and BC and then use the angle ACB and get the area using 1/2ab sinc
let me know if this helped😊
Original post by bluepotato
Alright so I have this question and I want to see how others have answered it.

The diagram shows triangle ABC with AC=8x-3, BC=4x-1, angle ABC=120 degrees and angle ACB=15 degrees.

(a) Show that the exact value of x is 9+√6/20 (7 marks)
(b) Find the area of triangle ABC giving your answer to 2 decimal places
(3 marks)


actually get the missing angle and then use sine rule and make sure u use the exact value of sin 45 and sin 120 to get to the answer they require u to get...
Original post by brainmaster
I'm seeing the first part u use the sine rule or cosine rule and then u simplify to make ur answer look like What They want then second one u just use replace the value of x in the equation for the length is AB and BC and then use the angle ACB and get the area using 1/2ab sinc
let me know if this helped😊


Could you do the question so I can see? We recently had this question in a test and I have to review it again after my teacher has marked it.
Original post by bluepotato
Could you do the question so I can see? We recently had this question in a test and I have to review it again after my teacher has marked it.


OK so you know two of the side lengths and all three of the angles (using the fact that the sum of the angles in a triangle must be 180 degrees).

So you need to put the information you have into the sine rule (i.e. sinA/A = sinB/B = sinC/C. This rule relies upon knowing pairs of angles and the lengths of their opposite sides. So the 120 degree angle is opposite the side of length (8x-3) and the 45 degree angle is opposite the length of side (4x-1). So now you have to work through the algebra of that. Remember that sin(180 - x) = sin(x) if you're not allowed a calculator!

The area of a triangle can be written as (1/2)*absinC so in this case that would be (1/2)*(4x-1)*(8x-3)*sin(15)
Original post by chailatte2407
OK so you know two of the side lengths and all three of the angles (using the fact that the sum of the angles in a triangle must be 180 degrees).

So you need to put the information you have into the sine rule (i.e. sinA/A = sinB/B = sinC/C. This rule relies upon knowing pairs of angles and the lengths of their opposite sides. So the 120 degree angle is opposite the side of length (8x-3) and the 45 degree angle is opposite the length of side (4x-1). So now you have to work through the algebra of that. Remember that sin(180 - x) = sin(x) if you're not allowed a calculator!

The area of a triangle can be written as (1/2)*absinC so in this case that would be (1/2)*(4x-1)*(8x-3)*sin(15)



Omds thanks so much!! I'll do it and try it out.
Original post by chailatte2407
OK so you know two of the side lengths and all three of the angles (using the fact that the sum of the angles in a triangle must be 180 degrees).

So you need to put the information you have into the sine rule (i.e. sinA/A = sinB/B = sinC/C. Thdidis rule relies upon knowing pairs of angles and the lengths of their opposite sides. So the 120 degree angle is opposite the side of length (8x-3) and the 45 degree angle is opposite the length of side (4x-1). So now you have to work through the algebra of that. Remember that sin(180 - x) = sin(x) if you're not allowed a calculator!

The area of a triangle can be written as (1/2)*absinC so in this case that would be (1/2)*(4x-1)*(8x-3)*sin(15)

did you get 0.26 for the area? it seems a bit small
Original post by desert_fairy
did you get 0.26 for the area? it seems a bit small

i was doing this question just now and came on here to check because i got that too but i also thought it was too small

if we both got it i'm assuming we've done it fine though?
Was it right?
I think it was 1.28 units^2
Reply 13
Yes, 0.26 is the area!
Original post by desert_fairy
did you get 0.26 for the area? it seems a bit small

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