Question

In the given figure , O is the center of the circle and \[\angle AOB={{110}^{\circ }}\]. Calculate \[\left( i \right)\angle ACO\] \[\left( ii \right)\angle CAO\]?

Answer 1

Hint: We will use geometry concepts to solve this problem. To find the given angles we should consider the \[\vartriangle AOC\]. First we will find any one angle in the triangle and then we will apply the theorems and rules to derive the unknown angles. So that we can find both the angles.

Complete step by step answer:
In this we know \[\angle AOB\] is \[{{110}^{\circ }}\].
As we all know the angle on a straight line is \[{{180}^{\circ }}\]. From the above figure we can say that
\[\angle AOB+\angle COA={{180}^{\circ }}\]
From this we can find \[\angle COA\] in \[\vartriangle AOC\]
Substituting the angle we have we will get
\[\Rightarrow {{110}^{\circ }}+\angle COA={{180}^{\circ }}\]
Now by substituting \[{{110}^{\circ }}\] from both sides we will get
\[\Rightarrow {{110}^{\circ }}+\angle COA-{{110}^{\circ }}={{180}^{\circ }}-{{110}^{\circ }}\]
By simplifying we get
\[\Rightarrow \angle COA={{70}^{\circ }}\]
So the angle \[\angle COA\] is \[{{70}^{\circ }}\].
In \[\vartriangle AOC\] now we know the angle \[\angle COA\] using this we will find \[\angle ACO\] and \[\angle CAO\].
We know that angle on the centre is twice the angle formed outside the center from the same segment.
In our diagram the angle on center is \[\angle AOB\] and the angle on the same segment is \[\angle ACO\].
So we can write them as
\[\angle AOB=2\times \angle ACO\]
Now we can divide both sides with 2. We will get
\[\Rightarrow \dfrac{1}{2}\times \angle AOB=\dfrac{2}{2}\times \angle ACO\]
By simplifying we will get
\[\Rightarrow \dfrac{1}{2}\times \angle AOB=\angle ACO\]
We know the angle \[\angle AOB\] so we substitute this in the above equation.
We have \[\angle AOB={{110}^{\circ }}\].
Substituting this we will get
\[\Rightarrow \dfrac{1}{2}\times {{110}^{\circ }}=\angle ACO\]
By simplifying we will get
\[\Rightarrow {{55}^{\circ }}=\angle ACO\]
By rewriting it we get
\[\Rightarrow \angle ACO={{55}^{\circ }}\]
Now we know both \[\angle ACO\] and \[\angle AOC\]so we can find \[\angle CAO\].
We all know that the sum of all angles in a triangle is equal to \[{{180}^{\circ }}\].
From this we can write
\[\Rightarrow \angle ACO+\angle CAO+\angle AOC={{180}^{\circ }}\]
Now substituting the values we have we will get
\[\Rightarrow {{55}^{\circ }}+\angle CAO+{{70}^{\circ }}={{180}^{\circ }}\]
Now by simplifying we will get
\[\Rightarrow \angle CAO+{{125}^{\circ }}={{180}^{\circ }}\]
Now subtracting \[{{125}^{\circ }}\] on both sides of the equation. we will get \[\Rightarrow \angle CAO+{{125}^{\circ }}-{{125}^{\circ }}={{180}^{\circ }}-{{125}^{\circ }}\]
By simplifying we get
\[\Rightarrow \angle CAO={{55}^{\circ }}\]
So we got the values of both the angles as
\[\angle ACO={{55}^{\circ }}\] and \[\angle CAO={{55}^{\circ }}\].

Note: we can also solve the angle \[CAO\] using radius values. Because the radius of the circle is equal and the angle formed by them is also equal. So we can write directly \[\angle CAO=\angle ACO\]. So that we can make it much simpler to solve.