Question

In the given figure, rays OA,OB,OC and OD are such that angle AOB= 56° angle BOC=100°,angle COD=x° and angle DOA = 74° find the value of x
sum of all the angles around a point is 360°.......

FRIENDS PLEASE ANSWER IT FAST MY EXAMS ARE GOING ON AND STEP BY STEP EXPLANATION PLEASE ​

Answer 1

Step-by-step explanation:

i have drawn a figure after reading your question.

and I have find the value of x°

as it is given that sum of all the angles around a point is 360° .

so with help of that , we get value of x°

x° = 130°

HOPE THIS HELPS

Answer 2

Given :

OA,OB,OC and OD are such that angle AOB= 56° angle BOC=100°,angle COD=x° and angle DOA = 74°

To find :

Value of x

Solution :

sum of all the angles around a point is 360°

=> ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

=> 56° + 100° + x + 74 = 360°

=> 156 + x + 74 = 360°

=> 230° + x = 360°

=> x = 360° - 230°

=> x = 130°

=> ∠COD = 130°

Sum other formula's :

• Volume of cylinder = πr²h

• T.S.A of cylinder = 2πrh + 2πr²

• Volume of cone = ⅓ πr²h

• C.S.A of cone = πrl

• T.S.A of cone = πrl + πr²

• Volume of cuboid = l × b × h

• C.S.A of cuboid = 2(l + b)h