Answer
466.2k+ views
Hint: Use the theorem “Angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference” and the exterior angle theorem of a triangle to find the desired angle in the equation.
Given that O is the center of the circle. Also, \[\angle COD={{40}^{o}}\] and \[\angle CBE={{100}^{o}}\].
Then, we have to find the measurement of \[\angle DAC\].
We know that the angle subtended by an arc of a circle at its center is twice the angle subtended anywhere on the circle’s circumference. (From theorem of the circle)
The proof to this theorem is quite simple and uses the exterior angle theorem – An exterior angle of a triangle is equal to the sum of the opposite interior angles.
Proof of theorem of a circle:
In the figure, we have an arc AC subtends to \[\angle AOB\] at the center O and \[\angle ACB\] at a point C on the circumference.
We have to prove that \[\angle AOB=2\times \angle ACB\]
Draw the line through O and C and let it intersect with the circle again at D.
From \[\Delta OAC\], we have
\[\angle OAC=\angle OCA....\left( i \right)\]
Because OA = OC, so the angle measure will be the same.
Similarly, in \[\Delta OBC\]we have
\[\angle OBC=\angle OCB.....\left( ii \right)\]
Because OB = OC the angle measure will be the same.
Thus from the equation (i) and (ii), we get
\[\Rightarrow \angle AOD=2\times \angle ACO.....\left( iii \right)\]
Twice because of the exterior angle theorem.
Also, \[\angle DDB=2\times \angle DCB....\left( iv \right)\]
Twice because of the exterior angle theorem.
Now, on adding equation (iii) and (iv), we have
\[\Rightarrow \angle AOD+\angle DOB=2\times \left( \angle ACO+\angle OCB \right)\]
\[\Rightarrow \angle AOB=2\times \angle ACB\]
As \[\angle AOD+\angle DOB=\angle AOB\]
Also, \[\left( \angle ACO+\angle OCB \right)=\angle ACB\]
Thus, \[\angle AOB=2\times \angle ACB\]
Or the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.
Using this theorem to find \[\angle DAC\] in our problem
In the figure, we can clearly see that \[\angle DOC\] and \[\angle DAC\], both lie on the same segment DC and also \[\angle DOC\] is at the center of the circle whereas \[\angle DAC\] is at the circumference of the circle.
Thus, from this we have
\[\angle DOC=2\times \angle DAC.....\left( v \right)\]
Given, \[\angle DOC={{40}^{o}}\]
Putting, \[\angle DOC=40\] in the equation (v), we get
\[\Rightarrow 40=2\times \angle DAC\]
Or, \[\Rightarrow \angle DAC=\dfrac{40}{2}\]
Or, \[\Rightarrow \angle DAC={{20}^{o}}\]
Because angle at any point on a circle by the same segment is half of the angle at the center.
Note: This problem can be solved in just a single step if the student knows that the angle subtended by an arc at the center of the circle is twice that of the angle subtended by the same arc at any point on the circumference of the circle. It is important for the students to know all the theorems to solve complex problems in very few steps.
Given that O is the center of the circle. Also, \[\angle COD={{40}^{o}}\] and \[\angle CBE={{100}^{o}}\].
Then, we have to find the measurement of \[\angle DAC\].
![seo images](https://www.vedantu.com/question-sets/1d5aa18f-ee93-41b2-b0d8-f0be119720d17308991756814977161.png)
We know that the angle subtended by an arc of a circle at its center is twice the angle subtended anywhere on the circle’s circumference. (From theorem of the circle)
The proof to this theorem is quite simple and uses the exterior angle theorem – An exterior angle of a triangle is equal to the sum of the opposite interior angles.
Proof of theorem of a circle:
![seo images](https://www.vedantu.com/question-sets/6de13746-1a06-4b5d-96dc-1f27e5bce22d388742371763700753.png)
In the figure, we have an arc AC subtends to \[\angle AOB\] at the center O and \[\angle ACB\] at a point C on the circumference.
We have to prove that \[\angle AOB=2\times \angle ACB\]
Draw the line through O and C and let it intersect with the circle again at D.
![seo images](https://www.vedantu.com/question-sets/1755caae-001f-466e-8aac-ab381a2d36831659353477325020846.png)
From \[\Delta OAC\], we have
\[\angle OAC=\angle OCA....\left( i \right)\]
Because OA = OC, so the angle measure will be the same.
Similarly, in \[\Delta OBC\]we have
\[\angle OBC=\angle OCB.....\left( ii \right)\]
Because OB = OC the angle measure will be the same.
Thus from the equation (i) and (ii), we get
\[\Rightarrow \angle AOD=2\times \angle ACO.....\left( iii \right)\]
Twice because of the exterior angle theorem.
Also, \[\angle DDB=2\times \angle DCB....\left( iv \right)\]
Twice because of the exterior angle theorem.
Now, on adding equation (iii) and (iv), we have
\[\Rightarrow \angle AOD+\angle DOB=2\times \left( \angle ACO+\angle OCB \right)\]
\[\Rightarrow \angle AOB=2\times \angle ACB\]
As \[\angle AOD+\angle DOB=\angle AOB\]
Also, \[\left( \angle ACO+\angle OCB \right)=\angle ACB\]
Thus, \[\angle AOB=2\times \angle ACB\]
Or the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.
Using this theorem to find \[\angle DAC\] in our problem
![seo images](https://www.vedantu.com/question-sets/93d919f6-1398-4dc5-804b-55c7de6602393824179107677508560.png)
In the figure, we can clearly see that \[\angle DOC\] and \[\angle DAC\], both lie on the same segment DC and also \[\angle DOC\] is at the center of the circle whereas \[\angle DAC\] is at the circumference of the circle.
Thus, from this we have
\[\angle DOC=2\times \angle DAC.....\left( v \right)\]
Given, \[\angle DOC={{40}^{o}}\]
Putting, \[\angle DOC=40\] in the equation (v), we get
\[\Rightarrow 40=2\times \angle DAC\]
Or, \[\Rightarrow \angle DAC=\dfrac{40}{2}\]
Or, \[\Rightarrow \angle DAC={{20}^{o}}\]
Because angle at any point on a circle by the same segment is half of the angle at the center.
Note: This problem can be solved in just a single step if the student knows that the angle subtended by an arc at the center of the circle is twice that of the angle subtended by the same arc at any point on the circumference of the circle. It is important for the students to know all the theorems to solve complex problems in very few steps.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)