Question
In the given figure, $AB || DE$ and $BD || FG$ such that $\angle\text{ABC}=50^\circ$ and $\angle\text{FGH}=120^\circ.$ Find the value of $x$ and $y$. 
✓
Answer
$\angle\text{FGH}+\angle\text{FGE}=180^\circ$ (linear pair)
$\Rightarrow120^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=60^\circ$ $AB || DF$ and $BD$ is the transversal.
$\Rightarrow\angle\text{CDE}=50^\circ$ $BD || FG$ and $DF$ is the transversal.
$\Rightarrow\angle\text{EFG}=\angle\text{CDE}$ (alternate angles)
$\Rightarrow\angle\text{EFD}=50^\circ$ In $\triangle\text{EFG},$ by angle sum property,
$\angle\text{FEG}+\angle\text{FGE}+\angle\text{EFG}=180^\circ$
$\Rightarrow\text{x}+\text{y}+50^\circ=180^\circ$
$\Rightarrow\text{x}+60^\circ+50^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
$\Rightarrow120^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=60^\circ$ $AB || DF$ and $BD$ is the transversal.
$\Rightarrow\angle\text{CDE}=50^\circ$ $BD || FG$ and $DF$ is the transversal.
$\Rightarrow\angle\text{EFG}=\angle\text{CDE}$ (alternate angles)
$\Rightarrow\angle\text{EFD}=50^\circ$ In $\triangle\text{EFG},$ by angle sum property,
$\angle\text{FEG}+\angle\text{FGE}+\angle\text{EFG}=180^\circ$
$\Rightarrow\text{x}+\text{y}+50^\circ=180^\circ$
$\Rightarrow\text{x}+60^\circ+50^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.
$BD$ is one of the diagonals of a quadrilaterl $ABCD$.
If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD},$show that
ar$($quadrilaterl $ABCD)$ $=\frac{1}{2}\times\text{BD}\times(\text{AL}+\text{CM}).$
→If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD},$show that
ar$($quadrilaterl $ABCD)$ $=\frac{1}{2}\times\text{BD}\times(\text{AL}+\text{CM}).$
In the adjoining figure, $ABCD$ is a trapezium in which $AB || DC; AB = 7\ cm; AD = BC = 5\ cm$ and the distance between $AB$ and $DC$ is $4\ cm$. Find the length of $DC$ and hence, find the area of trap. $ABCD$.
→If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
Question 15: Two lines $l$ and $m$ intersect at the point $0$ and $P$ is a point on a line $n$ passing through the point $0$ such that $P$ is equidistant from $l$ and $m.$ Prove that $n$ is the bisector of the angle formed by $l$ and $m.$
→In a $\triangle\text{ABC},$ $BM$ and $CN$ are perpendiculars from $B$ and $C$ respectively on any line passing through $A$. If $L$ is the mid-point of $BC$, prove that $ML = NL$.
→Two parallel sides of a trapezium are $60\ m$ and $77\ m$ and the other sides are $25\ m$ and $26\ m$. Find the area of the trapezium.
→Find the area of canvas required for a conical tent of height 24m and base radius $7m.$
→Show that $(x-2),(x+3)$ and $(x-4)$ are factors of $x^3-3 x^2-10 x+24$
→If $\text{a}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $\text{b}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}},$ show that $3\text{a}^2+4\text{ab}-3\text{b}^2=4+\frac{56}{3}\sqrt{10}.$
→