How to get the opposite angle?

$\begingroup$

I have been looking for the answer to this question but I haven’t found anything related. I will use the following image as an example.

Image

As you can see, there is a circle divided in different parts of the same size and their corresponding angle is indicated. So what do I mean with “getting the opposite angle”? Based on the image, let’s say I want to get the opposite angle of 60. While searching I have found that there’s one way to get the opposite angle, but it’s not exactly what I need. Those answers said that to get it I should add 180 to 60, but the problem is that it gives me 240, and that’s not what I need. I should get 120 as the value, the opposite horizontal angle. Is there any way to get this?

$\endgroup$ 1

3 Answers

$\begingroup$

It is the supplementary angle with a common sine value of the given angle.

$$ \sin x =\sin \alpha$$

$$ x_1= \alpha, x_2= 180^{\circ} -\alpha$$

It is not on the opposite side in the diagram you have given. It is mirrored with respect to y-axis.

$\endgroup$ $\begingroup$

You've got your concept wrong.. You don't add $180$ to $60$.. Take any angle say $x$. The opposite angle of $x$ when added with $x$ will always sum up to give 180.. So opposite angle of $x$ is equal to $(180-x)$.. Then what's gonna be the opposite angle of $60$? Exactly.. $180-60$.. which is $120$This is also known as the Linear pair property of angles.

$\endgroup$ 5 $\begingroup$

When you look at a pair of intersecting lines, you can see that four angles are formed around the point of intersection. The two angles that lie on the same line form an adjacent pair (summing to $180^{o} $), while the two that lie on different sides of the same line (facing each other) form a pair of opposite angles. With this in mind, had you named the points on the circle and its center, you could have been able to find it. The angles measure the same , then.

I said only what I know as a teen. Please let me know if this is correct.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like