Answer
424.8k+ views
Hint: We will solve this question by the sum of triangle. We will solve the question with the help of the angle sum property of a triangle which means the sum of all the angles in the triangle is equal to ${180^ \circ }$.
Complete step-by-step answer:
Let us first take the $\vartriangle PCE$
Angle 1 is exterior angle so, we use the property of exterior angle here
$\angle 1 = \angle b + \angle d$ …………..1
Now, in $\vartriangle BGD$ angle 2 is exterior angle for $\vartriangle BGD$
$\angle 2 = \angle c + \angle e$ ……………2
In $\vartriangle APG$ we know that sum of all the angle of the $\vartriangle APG$ is equal to ${180^ \circ }$
Angles in $\vartriangle APG$ are $\angle a, \angle 1 and \angle 2$
By angle sum property of a triangle
$\angle a + \angle 1 + \angle 2 = {180^{^ \circ }}$
Now, put the value of angle 1 and angle 2 by the equation 1 and 2
$\angle a + \angle d + \angle c + \angle e + \angle b = {180^ \circ }$ \[\]
So, the sum of the all the angles marked at the corners A,B,C,D,E is ${180^ \circ }$
So, the correct answer is “Option C”.
Note: In this question always remember that the angle of each exterior point is always the sum of two adjacent interior angles ${180^ \circ }$.
Alternative method: The adjoining star contains a regular pentagon, and we know the sum of interior angles is $ = {540^ \circ }$, and each angle is of the same measures so we can find the particular angle by dividing the sum by 5.
$ = \dfrac{{540}}{5}$
$ = {108^ \circ }$
It`s supplementary angles measure ${180^ \circ } - {108^ \circ } = {72^ \circ }$
The triangle formed by the star are isosceles triangle in which two angle are $ = {72^ \circ }$
So, we have the find the third angle
=${180^ \circ } - {72^ \circ } - {72^ \circ } = {36^ \circ }$
Each angle at the vertices of the star is equal to ${36^ \circ }$. Thus the sum of all the angle of the star
=${5 \times 36^ \circ } = {180^ \circ }$
Hence the answer is ${180^ \circ }$.
Option C is correct.
Complete step-by-step answer:
Let us first take the $\vartriangle PCE$
Angle 1 is exterior angle so, we use the property of exterior angle here
$\angle 1 = \angle b + \angle d$ …………..1
Now, in $\vartriangle BGD$ angle 2 is exterior angle for $\vartriangle BGD$
$\angle 2 = \angle c + \angle e$ ……………2
In $\vartriangle APG$ we know that sum of all the angle of the $\vartriangle APG$ is equal to ${180^ \circ }$
Angles in $\vartriangle APG$ are $\angle a, \angle 1 and \angle 2$
By angle sum property of a triangle
$\angle a + \angle 1 + \angle 2 = {180^{^ \circ }}$
Now, put the value of angle 1 and angle 2 by the equation 1 and 2
$\angle a + \angle d + \angle c + \angle e + \angle b = {180^ \circ }$ \[\]
So, the sum of the all the angles marked at the corners A,B,C,D,E is ${180^ \circ }$
So, the correct answer is “Option C”.
Note: In this question always remember that the angle of each exterior point is always the sum of two adjacent interior angles ${180^ \circ }$.
Alternative method: The adjoining star contains a regular pentagon, and we know the sum of interior angles is $ = {540^ \circ }$, and each angle is of the same measures so we can find the particular angle by dividing the sum by 5.
$ = \dfrac{{540}}{5}$
$ = {108^ \circ }$
It`s supplementary angles measure ${180^ \circ } - {108^ \circ } = {72^ \circ }$
The triangle formed by the star are isosceles triangle in which two angle are $ = {72^ \circ }$
So, we have the find the third angle
=${180^ \circ } - {72^ \circ } - {72^ \circ } = {36^ \circ }$
Each angle at the vertices of the star is equal to ${36^ \circ }$. Thus the sum of all the angle of the star
=${5 \times 36^ \circ } = {180^ \circ }$
Hence the answer is ${180^ \circ }$.
Option C is correct.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 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)
Give 10 examples for herbs , shrubs , climbers , creepers
![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)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english 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)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)