Question

In fig. 12, if AB || CD, CD || EF and y: z = 3:7, find​

Answer 1

Solution:

Solution:Given that y : z = 3 : 7

Solution:Given that y : z = 3 : 7Let ∠ y = 3a

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10a = 18

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10a = 18y = 3a = 3x18 = 54°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10a = 18y = 3a = 3x18 = 54°z = 7a = 7x18 = 126°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10a = 18y = 3a = 3x18 = 54°z = 7a = 7x18 = 126°x = z = 126°

Solution:Given that y : z = 3 : 7Let ∠ y = 3a Then ∠z = 7a ∠x and ∠z are alternate interior angles of parallel lines so that ∠x = ∠z …(1)Sum of interior angle on the same side of the transversal is always = 180°So that x + y = 180°plug the value of x from equation (1)z + y = 180°plug the value of z and y we get 7a + 3a = 180°10 a = 180°a = 180°/10a = 18y = 3a = 3x18 = 54°z = 7a = 7x18 = 126°x = z = 126°

Answer 2

Answer:

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