Math, asked by digvijayskashid, 6 months ago

 In the given below figure, side BC of ∆ABC has been produced to a point D. If ∠A = 3y⁰, ∠B = x⁰, ∠C = 5y⁰ and ∠CBD = 7y⁰. Then, the value of x is *

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Answered by abhimanyurp
217

Answer:

7y+5y=180(Linear pair)

12y=180

y=180/12

y=15

<ACB=5*15=75

<BAC=3*15=45

let <ABC=X

IN TRIANGLE ABC,

X+75+45=180 (ANGLE SUM PROPERTY)

X+120=180

X=180-120

X=60

Step-by-step explanation:

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Answered by amitnrw
16

Given :   In the given below figure, side BC of ∆ABC has been produced to a point D.

∠A = 3y⁰, ∠B = x⁰, ∠C = 5y⁰ and ∠CBD = 7y⁰.

To Find :  the value of x  

Solution:

7y⁰. and 5y⁰ forms a linear pair

Hence 7y⁰ +  5y⁰  = 180⁰

=> 12y⁰  = 180⁰

=> y = 15

Sum of angle of a triangle is 180⁰

3y⁰ +  5y⁰ + x⁰ = 180⁰

=> 8y⁰ + x⁰ = 180⁰

y = 15

=> 8(15°) + x⁰ = 180⁰

=> 120⁰ + x⁰ = 180⁰

=> x⁰ = 60⁰

=> x = 60

Value of x = 60

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