MHB Solving for $\theta$ and $H$ in Trig Equation

[Drain Brain]

I need help finding the unknowns

$H\cos(\theta)+559.68=750$
$H\sin(\theta)-124.26=0$

 

[Drain Brain]

I need help finding the unknowns

I know I have two unknowns and two equations but sin cos made me confuse when I try to solve one of the variable and plug it in one of the equations please help. Thanks!

$H\cos(\theta)+559.68=750$
$H\sin(\theta)-124.26=0$

 

[Fantini]

Hello. :) What have you tried?

 

[MarkFL]

I would first write the system in the form:

$$H\cos(\theta)=a$$

$$H\sin(\theta)=b$$

Square and add both equations, and you can eliminate $\theta$. What do you find?

edit: I have merged the two duplicate threads.

 

[Prove It]

I would first write the system in the form:

$$H\cos(\theta)=a$$

$$H\sin(\theta)=b$$

Square and add both equations, and you can eliminate $\theta$. What do you find?

edit: I have merged the two duplicate threads.

Or to avoid extraneous solutions, you can divide the second equation by the first, thereby making an equation just in terms of $\displaystyle \begin{align*} \tan{(\theta )} \end{align*}$...

 

[Drain Brain]

Hi MarkFL!

$\displaystyle H\cos(\theta)=a$(1)
$\displaystyle H\sin(\theta)=b$(2)

solvind for a and b

$a=190.32$
$b=124.26$

squaring and adding both equations I have

$H^2(\sin^2(\theta)+\cos^2(\theta))=(190.32)^2+(124.26)^2$
$H^2(1)= (190.32)^2+(124.26)^2$
$H=\sqrt((190.32)^2+(124.26)^2$
$H=227.29$

Solving for theta using equation (1) or (2)

Using (1)

$H\cos(\theta)=a$
$227.29\cos(\theta)=190.32$

$\theta=\cos^{-1}(\frac{190.32}{227.29})$
$\theta=33.14^{\circ}$

) hooray!

 

[MarkFL]

Does $H$ have to be positive?