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Help with triangle angles question watch

1. ..
2. (Original post by iamspiderman)

How would I start this question?
Start the question by using the sine rule to find angle BCA.

If you're still stuck, post all your working / ideas.
3. (Original post by iamspiderman)

How would I start this question?
Use the Sine Law.

sin(a) divided by side length A (opposite angle a), etc.
4. (Original post by iamspiderman)

How would I start this question?
We end up with:

6sin(45)=5sin(C)
sin(C) = 4243/5000
C=58.06 deg.

Angle B, therefore, has to be 76.94 deg.
5. (Original post by Schmeckel)
full solution
Please don't post full solutions - it's against the rules of this forum.

Your previous post was enough for the OP to get started and your solution was unnecessary.
6. ..
7. (Original post by iamspiderman)
Well I basically got the opposite of what you got.
I did
Sin45/5 = sinC /6

How are you meant to know which angle goes with which side ?

Ok thanks
Oh, I cross-multiplied to get rid of all those messy fractions.
6*(sin45/5)=(sinC/6)*5

That makes them both fractions with 30 as the denominator and then you can cancel them out.

Angle A is directly opposite side a. Angle B is directly opposite side b. Angle C is directly opposite side c.
8. (Original post by iamspiderman)
Well I basically got the opposite of what you got.
I did
Sin45/5 = sinC /6

How are you meant to know which angle goes with which side ?
Ok thanks
If you rearrange that you get:

6sin(45) = 5sin(C)

which is the same as schmeckel's solution.

From your line, you should multiply both sides by 6 to get

sin(C) = 6sin(45)/5

Do you know what to do next?
9. (Original post by Schmeckel)
Angle B, therefore, has to be 76.94 deg.
As well as the complete solution issue that has been mentioned

The OP was not looking for angle B they are asked for angle C

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