Help with a simultaneous trig problem

[ScottH]

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 !

 

[tiny-tim]

Welcome to PF!

Hi ScottH! Welcome to PF!

(have a theta: θ and a phi: φ and a degree: ° )

Here's two tricks you need to become familiar with:

i] Asinθ + Bcosθ cab be written as Csin(θ + φ), where tanφ = B/A

ii] eliminate θ first (instead of b) by using cos²θ + sin²θ = 1

 

[ScottH]

Hi there tiny-tim. Thanks for your reply. I'm no mathematician by any stretch, just trying to get my head around this as part of my engineering course. Could you possibly show me how you would use these tricks in my equation? If you can that would be great. Again, thanks for your time in replying.
Cheers ScottH

 

[tiny-tim]

Hi ScottH!

Sorry, on this forum you have to do the work yourself.

Start with your 1375 cosθ tan 30 + 1375 sinθ …

put it in the from Csin(θ + φ) ​

 

[ScottH]

Thanks tiny tim- I'll see how I go. Have a great day !