Question

how do you solve for the unknown angles in these two equations?

-20cos(alpha)-30cos(theta)+40=0 eq.1
+20sin(alpha)-30sin(theta) = 0 eq.2

Answer 1

help :)

Answer 2

to start with you can divide both equations by 10 to simplify them

Answer 3

then use the eq 1 to get and expression for cos(alpha) and eq 2 to get and expression for sin(alpha)

Answer 4

finally use the identity:\[\cos^2(\alpha)+\sin^2(\alpha)=1\]to solve this

Answer 5

please, note that I can write your equations as below:
\[2 \cos \alpha +3 \cos \theta =4,2 \sin \alpha-3 \sin \theta=0\]
now, please square both sides of those equation, and then add them side by side:

Answer 6

i'm trying to solve it :)

Answer 7

after my steps, you should get this condition:
\[\cos (\alpha+\theta)=\frac{ 1 }{ 2 }\]
please check!

Answer 8

I got cos(alpha+theta) = 1/4

Answer 9

that's right! I have made an error

Answer 10

@moongazer - have you tried the method I suggested?

Answer 11

the question is asking you to find both alpha and theta

Answer 12

@asnaseer not yet. I'll try it too

Answer 13

now we have a relation between alpha and theta, which we can use together other two equations

Answer 14

\(\large\tt \begin{align} \color{black}{-20cos(\alpha)-30cos(\theta)+40=0 \\
20sin(\alpha)-30sin(\theta) = 0\\~\\~\\~\\
2cos(\alpha)+3cos(\theta)=4 \\~\\
2sin(\alpha)-3sin(\theta) = 0\\~\\~\\~\\
2cos(\alpha)=4-3cos(\theta)\\~\\
2sin(\alpha) = 3sin(\theta)\\~\\~\\~\\
\text{squaring}\\~\\~\\~\\
4cos^2(\alpha)=16-24cos(\theta)+9cos^2(\theta)---------\color{red}{(1)}\\~\\
4sin^2(\alpha) =9sin^2(\theta)---------\color{red}{(2)}\\~\\}\end{align}\)

add 1 and 2 this should give the answr quickly

Answer 15

multipling by sin theta your first equation, we get:
\[2 \sin \alpha \sin \theta =3(\sin \theta)^{2}\]
whereas multipling both sides of your second equation by cos theta, we find:
\[2 \cos \alpha \cos \theta=4 \cos \theta -3(\cos \theta)\]

Answer 16

I got the answer doing what asnaseer said: )
the answer is
alpha=46.567
theta=28.955
thanks :)

Answer 17

oops...\[2 \cos \alpha \cos \theta =4 \cos \theta -3 (\cos \theta)^{2}\].

Answer 18

@Michele_Laino I'll also try your solution. :) thanks

Answer 19

@moongazer - yw :)

Answer 20

substituting those above relation into the condition cos(Alpha+theta)=1/4, we get:
\[3-4 \cos \theta =\frac{ 1 }{ 4 }\]

Answer 21

Thanks! @moongazer

Answer 22

oops...\[3-4 \cos \theta=\frac{ 1 }{ 2 }\]

Answer 23

@Michele_Laino I think it should be:
1/2 = 4cos(theta) - 3

I also got the answer using your method. :)
thanks!

Answer 24

yes! you are right, I'm very sorry!

Answer 25

thanks! again! @moongazer