Lines and Angles — Question 1

It is given that y : z = 3 : 7

⇒ $\dfrac{y}{z} = \dfrac{3}{7}$

⇒ y = $\dfrac{3z}{7}$ ........(1)

Let ∠CON = p

Since, CD || EF

∴ p = z (Alternate interior angles are equal) .....(2)

⇒ y + p = 180° (Linear pairs)

⇒ y + z = 180° [From (2)]

Substituting value of y from equation (1), in above equation, we get :

$\Rightarrow \dfrac{3z}{7} + z = 180° \\[1em] \Rightarrow \dfrac{3z + 7z}{7} = 180° \\[1em] \Rightarrow \dfrac{10z}{7} = 180° \\[1em] \Rightarrow z = \dfrac{7}{10} \times 180° \\[1em] \Rightarrow z = 126° \\[1em] \Rightarrow y = \dfrac{3}{7}z \\[1em] \Rightarrow y = \dfrac{3}{7} \times 126° \\[1em] \Rightarrow y = 3 \times 18° \\[1em] \Rightarrow y = 54°.$

From figure,

AB || CD & MN is transversal

We know that, the sum of co-interior angles are supplementary.

⇒ x + y = 180°

⇒ x + 54° = 180°

⇒ x = 180° - 54°

⇒ x = 126°

Hence, x = 126°.