Calculate the value of ##θ## and ##X##

[chwala]

Homework Statement

See attached- i am doing self learning in this area...insight is welcome

Relevant Equations

Mechanics

My take,

$5 \cos 0 = 10 \cos θ$

$\cos θ = 0.5$

$⇒θ = 60^0$

and

$X= 10 \cos (90^0-θ)=\cos 30^0= 8.66$ (to two decimal places).

...insight welcome

 

[WWGD]

Not sure anyone works with degrees outside a classroom anymore. From a Mathematical perspective, it may be a good idea to bring up the periodicity of the (arc)sine.

 

[Hill]

Looks good. Just one small inconsistency in your equations:
$5 \cos 0 = 10 \cos θ$
but
$X= 10 \cos 30^0$.
Why the difference?

 

[chwala]

Looks good. Just one small inconsistency in your equations:
$5 \cos 0 = 10 \cos θ$
but
$X= 10 \cos 30^0$.
Why the difference?

Let me edit that...

 

[Bosko]

My take,

$5 \cos 0 = 10 \cos θ$

$\cos θ = 0.5$

$⇒θ = 60^0$

This is correct
"cos 0" is redundant. You can start with $5 = 10 \cos \theta$

and

$X= 10 \cos 30^0= 8.66$ (to two decimal places).

...insight welcome

.. and this. All looks fine.
Instead $\cos 30^o$ I would put $\sin \theta$ . ( $\theta=60^o$ )
Certainly, both are the same number $\sqrt{3}/2.$

 

[kuruman]

Since you're learning and for future reference.

When $N$ vectors add to zero, they form a closed $N$-sided polygon, in this case a triangle (figure on the right drawn to scale). Because it is a right triangle, you can find the unknown side by using the Pythagorean theorem $$X=\sqrt{10^2-5^2}~\text{N}.$$Then $$\cos\theta=\frac{5}{10}\implies \theta=60^{\circ}.$$

 

[chwala]

View attachment 340154Since you're learning and for future reference.

When $N$ vectors add to zero, they form a closed $N$-sided polygon, in this case a triangle (figure on the right drawn to scale). Because it is a right triangle, you can find the unknown side by using the Pythagorean theorem $$X=\sqrt{10^2-5^2}~\text{N}.$$Then $$\cos\theta=\frac{5}{10}\implies \theta=60^{\circ}.$$

Nice, i can see that if the $10$ N line is extended to the first quadrant and noting that the angle $θ$ is vertically opposite then we can use your approach.