Question

Help explain how to solve simultaneously the below equations:

10 = v cos theta + 5.638
4 = v sin theta - 2.05

I need to figure out angle theta and v?

Answer 1

I need to know what to solve for in the 1st equation? I know it seems easy, I'm confusing myself a bit too much. Help!

Answer 2

First you seperate the vcos thete & vsin theta & take their squre.
After squaring add them to find v & the put the value of v in any equation you can get theta...

Answer 3

A fellow user helped me with this: v cos theta = 4.362 and v sin theta = 6.05

I divide (v sin theta/v cos theta) = (6.05/4.362)
v tan theta = 1.39
inverse tan of 1.39 = 54.21 degrees
I substituted that into 1st equations = v (cos 54.21) = 4.362
v = 7.46 m/s

Answer 4

\[10=v \cos \Theta+5.638\]
\[\rightarrow v \cos \Theta =10-5.638\]
\[\rightarrow v ^{2} \cos ^{2} \Theta=(4.362)^{2}\]
\[\rightarrow v ^{2} \cos ^{2} \theta=19.027..................1 no.\]

and then

\[4=v \sin \Theta-2.05\]
\[\rightarrow v ^{2} \sin ^{2} \Theta=(4+2.05)^{2}\]
\[\rightarrow v ^{2} \sin ^{2} \Theta=36.603..................2 no\]


then add 1no. and 2.no

\[v^{2}\sin^{2}\Theta+v^{2}\cos^{2}\theta=19.027+36.603\]
\[v^{2}=55.63\]
\[v=\sqrt{55.63}\]
\[v=7.459...(Answer)\]

Answer 5

@biswajit_paul where is sin2theta and cos2theta go when we added them?

Answer 6

is it a trig rule?

Answer 7

@Boushra It comes from euclidean geometry.
\[\sin^{2}\theta + \cos^{2}\theta=1\]

Answer 8

It is trigonometry

Answer 9

super! I will keep that in mind :) thank you so much!