In the given figure,if PQ || RS,RS || MN and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the transversal line,but looking at the positions,$x$ and $z$ are interior angles on the same side of the transversal for lines $PQ$ and $MN$,so $x + z = 180^o$.Also,$y$ and $z$ are consecutive interior angles between lines $RS$ and $MN$,so $y + z = 180^o$.This implies $y = x$ (as both are supplementary to $z$).Given $y: z = 7: 11$,let $y = 7k$ and $z = 11k$.Since $y + z = 180^o$,we have $7k + 11k = 180^o$,which gives $18k = 180^o$,so $k = 10^o$.Thus,$y = 7 	imes 10^o = 70^o$ and $z = 11 	imes 10^o = 110^o$.Since $x + z = 180^o$ (consecutive interior angles for $PQ || MN$),we have $x + 110^o = 180^o$.Therefore,$x = 180^o - 110^o = 70^o$.Wait,checking the figure again: $x$ and $y$ are consecutive interior angles between $PQ$ and $RS$,so $x + y = 180^o$. Since $y = 70^o$,$x = 180^o - 70^o = 110^o$.

In the given figure,if $PQ || RS$,$RS || MN$ and $y: z = 7: 11$,then find $x$. (in $^o$)

Similar Questions

In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $I$. Prove that $\angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.

$A$ triangle $ABC$ is right-angled at $A$. $L$ is a point on $BC$ such that $AL \perp BC$. Prove that $\angle BAL = \angle ACB$.

In the figure,$AB$,$CD$,and $EF$ are three lines concurrent at $O$. Find the value of $y$. (in $^{\circ}$)

In the figure,$DE \parallel QR$ and $AP$ and $BP$ are bisectors of $\angle EAB$ and $\angle RBA$ respectively. Find $\angle APB.$ (in $^{\circ}$)

In the figure,$m$ and $n$ are two plane mirrors perpendicular to each other. Show that the incident ray $CA$ is parallel to the reflected ray $BD$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module