# Comparing angles opposite radius of a circle

• zak100
In summary, the homework statement states that BC is greater than CD. However, the attempt at a solution reveals that both triangles are isosceles, meaning that x and y should be equal because they are both opposite the radius. However, the book states a different answer, stating that x is greater than y.
zak100

## Homework Statement

In circle O , BC >CD
Compare x & y (which is greater?)

## Homework Equations

There are no eq. Rule: angle opposite larger side is larger

## The Attempt at a Solution

In my view both triangles are isosceles triangle. So x & y should be equal because they are both opposite the radius

Zulfi.

zak100 said:
because they are both opposite the radius

Why ?
Imagine a 6,8,10 right triangle and a equilateral triangle with side 6. Is in both cases angle opposite to side 6 same as 60° (use trigonometry if you don't have the time to construct the triangles).

Last edited:
Hi,
<Why ?>
OC is the radius of circle.
<6,8,10 right triangle>
In this case, the angle opposite to 6 would be the smallest.
<equilateral triangle>
In this case angle should be 60 & sides must be equal not necessary 6.

Zulfi.

zak100 said:
OC is the radius of circle.

Yes, but does that make the angle x and y equal ?

zak100 said:
<6,8,10 right triangle>
In this case, the angle opposite to 6 would be the smallest.
<equilateral triangle>
In this case angle should be 60 & sides must be equal not necessary 6.

Yes you are correct.
So what do we deduce from this ?
_______________________________________________________
Also compare ##\angle BOC##and ##\angle COD##, which is greater ?

Last edited:
Since its given that BC is greater therefore, arc BC is greater so angle BOC> DOC. But I don't know about x & y? Only that they are opposite to radius so they must be equal.

Zulfi.

zak100 said:
so angle BOC> DOC.
Right. So can you write x in terms of angle BOC and y in terms of DOC?

zak100 said:
Only that they are opposite to radius so they must be equal.
There is no such rule. E.g. consider B starting near C and moving around to be almost opposite C. At first, x is nearly 90o; half way round it is 45o; when opposite it is 0.
The angle subtended by a diameter is always a right angle. Maybe you are thinking of that.

Hi,
Thanks for your example. But i can't understand it. But i want to give you an example of isosceles triangle in which angles opposite to the equal sides are equal and the triangle formed by connecting two radii is an isosceles triangle.
<So can you write x in terms of angle BOC and y in terms of DOC?>
Based upon my understanding:
x + <BOC +<BCO = 180
or <BOC= 180 -2x
&
<DOC = 180 -2y
Kindly guide me.
Zulfi.

zak100 said:
<BOC= 180 -2x
&
<DOC = 180 -2y
Correct.
So x=? and y=? (in terms of BOC and DOC). If x and y were equal, BOC and DOC would also be equal. But BOC>DOC, meaning x and y are not equal. Which one is greater then?

zak100 said:
x + <BOC +<BCO = 180
or <BOC= 180 -2x
&
<DOC = 180 -2y

$$x = {180- \angle BOC \over 2}$$
$$y = {180- \angle DOC \over 2}$$

Now which is bigger ?

zak100 said:
But i want to give you an example of isosceles triangle in which angles opposite to the equal sides are equal and the triangle formed by connecting two radii is an isosceles triangle.

Your case is true for two angles opposite to equal sides in a triangle. Here we are talking about two different triangles with different bases. So your rule will not work here.

Hi,
From these equation it looks that y is bigger. But you have derived this eq based upon my work and i used the logic that ∠x is equal to ∠BCO because BOC is an isosceles triangle and ∠x is opposite to OC. Similarly ∠y is opposite to OC, so its astonishing that ∠x & ∠y are not equal?

Zulfi.

zak100 said:
BOC is an isosceles triangle and ∠x is opposite to OC. Similarly ∠y is opposite to OC, s
Yes, BOC is isosceles, with OB=OC, so angle OBC=angle OCB.
Likewise, angle OCD=angle ODC.
But these are two different isosceles triangles; there is no rule that says they must have the same angles as each other, even if they have the same lengths on some sides.
Simple experiment: draw a triangle with sides length 4, 4, 7 and another with sides length 4, 4, 1. Each has two equal angles, but that angle is very different in one from the other.

## 1. What is the relationship between the angles opposite the radius of a circle?

The angles opposite the radius of a circle are equal. This is because a radius is a straight line that connects the center of the circle to any point on the circumference, and it bisects the angle formed by two tangent lines at that point.

## 2. How do you compare angles opposite the radius of a circle?

To compare angles opposite the radius of a circle, you can measure the angles using a protractor or use the formula angle = 180 - (360 / number of sides) if the circle is divided into equal parts.

## 3. Can the angles opposite the radius of a circle be different?

No, the angles opposite the radius of a circle are always equal. This is a property of circles and is true for all circles, regardless of their size.

## 4. Why are the angles opposite the radius of a circle important?

The angles opposite the radius of a circle are important because they help us to understand the relationship between the radius and the circumference of a circle. They also help us to solve problems involving circles, such as finding the area or perimeter.

## 5. Is there a real-life application for comparing angles opposite the radius of a circle?

Yes, there are many real-life applications for comparing angles opposite the radius of a circle. For example, in geometry, this concept is used to calculate the circumference and area of a circle. In engineering, it is used to design and construct circular structures such as bridges and tunnels. In navigation, it is used for plotting and calculating the distance between points on a map.

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