Exercise 10.2
Question 1
(i) $\angle 2, \angle 6 \rightarrow$ Corresponding angles
(ii) $\angle 1, \angle 6 \rightarrow$
(iii) $\angle 3, \angle 5 \rightarrow$ Alternative interior angles
(iv) $\angle 2, \angle 7 \rightarrow$
(v) $\angle 3, \angle 6 \rightarrow$ Co-interior angles
(vi) $\angle 4,<8 \rightarrow$ Corresponding angles
Question 2
(i) If a transverse line cuts two parallel lines , Then corresponding angles are equal.
(ii) If two lines are cut by a transversal line such that a present of Alternative angles are equal , then the lines are parallel
(iii) If two lines are cut by a transversal line such that a pair of co-interior angles are supplementary then the lines are parallel.
Question 3
(i) $\quad x=100^{\circ} \quad(\therefore$ Corresponding angles $)$
(ii) $x+110^{\circ}=180$ (∵pair of co- interior angles are supplementary)
$x=180-110$
$x=70^{\circ}$
(iii) (diagram to be added)
From figure
$x+110^{\circ}=180^{\circ}$ (∵Froms a linear pair)
x= 180 - 110
x = $70^{\circ}$
Question 4
(i) From the figure
Given angles 2x+ 6, 3x + $54^{\circ}$ are Co- interior angles
When pair of parallel lines Cut by transversal line , of co- interior angles are supplementary
$\begin{aligned} 2 x+6+3 x+54 &=180^{\circ} \\ 5 x+60 &=180 \\ 5 x &=180-60 \\ 5 x &=120 \\ x &=\frac{120}{5} \\ & x=24^{\circ} \end{aligned}$
(ii) (diagram to be added)
From the figure
$2 x+15^{\circ}, \quad 3 x+30^{\circ}$ are co- interior angles
∴ Co- interior angles are supplementary
$\begin{array}{rl}2 x+15+3 x+30 & =180 \\ 5 x+45 & =180 \\ 5 x & =180-45 \\ 5 x & =135 \\ 6 & x=\frac{135}{5} \\ \ & x=27^{\circ}\end{array}$
Question 5
(i) From given figure
x $=60^{\circ}$ (∵vertically opposite angles)
$x=y=60^{\circ}$ (∵Corresponding angles)
(ii) From given figure
q= $135^{\circ}$ (∵Vertically opposite angles)
q + p = $180^{\circ}$ (∵a pair of co-interior angles are supplementary)
$\begin{aligned} 135+P &=180^{\circ} \\ P &=180-135 \\ P &=45^{\circ} \end{aligned}$
(iii) From given figure
a $=70^{\circ}$ (∵ Alternative interior angles are equal)
$a+b=180^{\circ}$ (∵ Forms a linear pair)
$\begin{aligned} 70+b &=180 \\ b &=180-70 \\ b &=110^{\circ} \end{aligned}$
(iv) From given figure
Z= $128^{\circ} (∵Corresponding angles are equal)
$x+128^{\circ}=180 (∵Forms a linear pair)
$x=180-128^{\circ}$
$x=52^{\circ}$
$x=y=52^{\circ}$ (∵ Corresponding angles are equal)
$\therefore \quad x=52^{\circ}, y=52^{\circ}, \quad z=128^{\circ}$
(v) From given figure
b = $75^{\circ}$ $(\therefore$ vertically opposite angles)
$a+75^{\circ}=180^{\circ}$ (∵Forms a linear pair)
$a=180-75$
$a=105^{\circ}$
C = $75^{\circ}$ (∵Corresponding angles are equal)
$\begin{aligned} d+C=180^{\circ} \\ d+75 &=180 \\ d &=180-75 \\ d &=105^{\circ} \end{aligned}$
(vi) From given figure
p = $62^{\circ}$ (∵Vertically opposite angles )
q + p = $180^{\circ}$ (∵Forms a linear pair)
$\begin{aligned} q+62 &=180 \\ & q=180-62 \\ & q=118^{\circ} \end{aligned}$
$p=S=62^{\circ}$ (∵Corresponding angles are equal)
$q = r = 180^{\circ}$ (∵ Alternative interior angles )
Question 6
( diagram to be added)
Consider $E F \| C D, E C$ transversal line
ஃ $x+120=180^{\circ}$ (∵Pair of Co-interior angles are supplementary)
$x=180-120$
$1 x=60^{\circ}$
Consider $C D \| A B, A C$ Transversal line
$y+140=180 \quad$ (\because$ pair of co- interior angles are Supplementary)
$\angle E C A=x+y=60+40=100^{\circ}$
Question 7
( diagram to be added)
$l \| m, A C$ is transversal line
$\therefore Z=55^{\circ}$ (∵ Alternative interior angles )
$l \| m, A B is transversal line
x = $45^{\circ}$ (∵ Alternative interior angles)
In triangle ABC
$\begin{aligned} 45+55+y &=180 \\ y+100 &=180 \\ y &=180-100 \\ y &=80^{\circ} \end{aligned}$
Question 8
( diagram to be added)
$l \| m, A C$ transversal line
z+$60^{\circ}=180^{\circ}$ (∵Pair of Co -interior angles are supplementary)
$z=180-60$
$z=120^{\circ}$
$x+y=60^{\circ}$-------① (∵Alternative interior angles are equal)
l|lm, $A B$ transversal line
$y+z=145^{\circ}$ (∵Alternative interior angles are equal)
$y+120=143^{\circ}$
$y= 23^{\circ}$
Substituted y value is equation ①
$x+23=60$
$x=60-23$
$x=37^{\circ}$
(ii) ( diagram to be added)
$l \| m, BC$ is transversal line
$a=55^{\circ}$ (∵Corresponding angles are equal)
In triangle ABC
$\begin{aligned} 72+a+b=& 180^{\circ} \\ 72-55+b=& 180^{\circ} \\ b+127=180^{\circ} \\ b=180-127 \\ b=53^{\circ} \end{aligned}$
$b+c+55=180^{\circ}$ (Forms a linear pair)
$\begin{aligned} 53+c+55 &=180^{\circ} \\ c+108 &=180 \\ C &=180-108 \\ C &=72^{\circ} \end{aligned}$
(iii) ( diagram to be added)
From the given figure
$75+a=180^{\circ}$ (∵Forma linear pair)
$a=180-75$
$a=105^{\circ}$
$l \| m$, q is transversal line
b = $75^{\circ}$ (∵Alternative interior angles are equal)
$P \| q$, $l$ is transversal line
d = b = $75^{\circ}$ (∵Corresponding angles are equal)
$l \| m$ p is transversal line
d = c = $75^{\circ}$ (∵Alternative interior angles are equal)
Question 9
( diagram to be added)
(i) From the given figure
106, 64 are co- interior angles
Sum = $106+64=170^{\circ} \neq 180^{\circ}$
$\therefore \quad l,m are not parllel
(ii) ( diagram to be added)
From the given figure
Pair of co- interior angles $75^{\circ,75^{\circ}$
Sum= $75+75=150 \Rightarrow 180^{\circ}$
∴ l, m are not parallel
(iii) ( diagram to be added)
Pair of co-interior angles are $57^{\circ},123^{\circ}$
Sum =$123^{\circ}-157^{\circ}=180$
∴ l, m are parallel to each other.
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