- #1
ScottH
- 3
- 0
Hi all. I have to solve this simultaneous trig equation and I am stumped. Can anybody give me a worked solution please?
b*cos 30 = 1375*cos theta (i)
b*sin 30 + 1375 sin theta = 1500 (ii)
I end up with:
b= 1375cos theta / cos 30
substituting into (ii)
1375 cos theta/cos30 * sin 30 + 1375 sin theta = 1500
1375 cos theta * sin30/cos30 + 1375 sin theta = 1500
1375 cos theta tan 30 + 1375 sin theta = 1500
From here I'm lost
Please help !
b*cos 30 = 1375*cos theta (i)
b*sin 30 + 1375 sin theta = 1500 (ii)
I end up with:
b= 1375cos theta / cos 30
substituting into (ii)
1375 cos theta/cos30 * sin 30 + 1375 sin theta = 1500
1375 cos theta * sin30/cos30 + 1375 sin theta = 1500
1375 cos theta tan 30 + 1375 sin theta = 1500
From here I'm lost
Please help !