If you take Tan inverse of 7/10 you get the angle at A = 34.992 Deg. Then use the Sine rule as follows: √61.25/sin34.992 =6.5/sinC. When you solve for C you get an angle of 28.44 Deg. I wouldn't worry about it at this stage. Sounds like you got most of it correct.
Malloy maths is the goat
Thanks.
🐐
can you do q4 part b with the sine rule?
You can if you had one of the other angles. You can work out the angle at B using Trig ratios but I didn't go into that method.
MolloyMaths yeah, I did it with 2 sides and an angle. I got CB using Pythagoras's theorem. I think I got 30 something degrees though as the answer.
If you take Tan inverse of 7/10 you get the angle at A = 34.992 Deg. Then use the Sine rule as follows:
√61.25/sin34.992 =6.5/sinC. When you solve for C you get an angle of 28.44 Deg.
I wouldn't worry about it at this stage. Sounds like you got most of it correct.
Thanks for the comment on Q8 Paper 2 2017. Well spotted. I've redone this video with z = -1.35
MolloyMaths no problem, in Part 3 of q8 Could I use the x+/- 1.96(standard error) formula getting values of 61 and 63
nice one