Exercise 10.1
Question 1
(i) No
(ii) yes
(iii)yes
(iv) Yes
(v) Yes
(vii) yes
Question 2
(i) Let x be Complement angle for 25
$\begin{aligned} \therefore x+25 &=90 \\ x &=90-25 \\ x &=65^{\circ} \end{aligned}$
$\therefore 65^{\circ}$ is complementary angle for $25^{\circ}$
(ii) let $x$ be complement angle for $6.3^{\circ}$
$\begin{aligned} x+63 &=90 \\ x &=90-63 \\ x &=27 \end{aligned}$
∴ $27^{\circ}$ in Complementary angle for $63^{\circ}$
iii) Let x be complementary angle for given angle
∴ $\begin{aligned} x+57^{\circ} &=90 \\ x &=90-57 \\ x &=33^{\circ} \end{aligned}$
∴ $33^{\circ}$ in Complementary angle for $57^{\circ}$
Question 3
Given angle $=105^{\circ}$
Let 'x' be supplementary angle for given angle
∴ $\begin{aligned} x+105 &=180 \\ x &=180-105 \\ x &=75 \end{aligned}$
∴ $75^{\circ}$ in Supplementary angle for $105^{\circ}$
Question 4
(i) Given angle $=87^{\circ}$
Let 'x' be supplementary angle for $87^{\circ}$
$\begin{aligned} \therefore \quad x+87 &=180 \\ x &=180-87 \\ x &=93^{\circ} \end{aligned}$
$\therefore 93^{\circ}$ is supplementary angle for $87^{\circ}$
(ii) Given angle $87^{\circ}$
Let x be the supplement to the $87^{\circ}$
$\begin{aligned} \therefore \quad x+87^{\circ} &=180 \\ x &=180-87 \\ x &=93^{\circ} \end{aligned}$
(iii) Given angle = 142$^{\circ}$
Let x be the supplement to 142$^{\circ}$
$\begin{aligned} \therefore x+142 &=180^{\circ} \\ x &=180-142 \\ x &=38^{\circ} \end{aligned}$
∴ $38^{\circ}$ is supplement to 142$^{\circ}$
Q.4
(i) $55^{\circ}, 125^{\circ}$
Given angles $=55^{\circ}, 125^{\circ}$
sum $=55+125=180^{\circ}$
∴ Given angles of supplementary
(ii) $34^{\circ}, 56^{\circ}$
Given angles' $=34^{\circ}, 56^{\circ}$
sum $=34+56=90^{\circ}$
∴ Given angles are Complementary.
(iii) $137^{\circ}, 43^{\circ}$
Given angles = $137^{\circ}, 43^{\circ}$
Sum $=137+43=180^{\circ}$
∴ Given angles of supplementary
(iv) $112^{\circ}, 68^{\circ}$
Given angles $=112^{\circ}, 68^{\circ}$
Sum $=112^{\circ}+68=180^{\circ}$
∴ Given angles are Supplementary
(v) $45^{\circ}, 45^{\circ}$
Given angles = $45 ,45^{\circ}$
Sum $=45+45=90^{\circ}$
∴ Given angles are Complementary.
(vi) $72^{\circ}, 18^{\circ}$
Given angles $=72^{\circ}, 18^{\circ}$
Sum $=72+18=90^{\circ}$
∴ Given angles are Complementary.
Question 5
(i) Let 'x' be the angle and its complement is also 'x '
$\begin{aligned} \therefore \quad x+x &=90^{\circ} \\ 2 x &=90^{\circ} \\ x &=45^{\circ} \end{aligned}$
Hence, required angle = $45^{\circ}$
$45^{\circ}$ is the angle which equal to its complement
(ii) Let 'x' be the angle and it's supplement is also 'x '
$\begin{aligned} \therefore x+x &=180^{\circ} \\ 2 x &=180^{\circ} \\ x &=\frac{180}{2} \\ x &=90 \end{aligned}$
∴ $90^{\circ}$ is the angle which equal to its supplement
Question 6
Given Complementary angles = $(x+4)^{\circ},(2 x-7)^{\circ}$
$\begin{array}{rl}\therefore \quad x+4+2 x-7 & =90 \\ 3 x-3 & =90 \\ 3 x & 290+3 \\ 3 x & 293 \\ x & =\frac{93}{3} \\ x & =31^{\circ}\end{array}$
Question 7
Given supplementary angles ratio = 2:7
Let the supplementary angles = 2x, 7x
$\begin{aligned} \therefore \quad 2 x+7 x &=180 \\ 9 x &=180 \\ x &=\frac{180}{9} \\ & x=20 \end{aligned}$
∴ The given angles = $2 x=2 \times 20=40^{\circ}$
$7 x=7 \times 20=140^{\circ}$
Question 8
Let the Smaller angle be = $x^{\circ}$
Larger angles = $x^{\circ}+44^{\circ}$
$x^{\circ}, x+44^{\circ}$ are supplementary angles
$\begin{aligned} \therefore x+x+44=& 180 \\ 2 x+44 &=180^{\circ} \\ 2 x &=136 \\ x &=\frac{136}{2} \\ x &=68^{\circ} \end{aligned}$
∴ Angles = x = $68^{\circ}
$x+44=68+44=112^{\circ}$
∴ Supplementary angles = $68^{\circ},112^{\circ}$
Question 9
Let 'x' be an angles
Given x equal to half of its complement
Complement angle to 'x ' = 90 - x
$\begin{aligned} \therefore x=& \frac{90-x}{2} \\ 2 x &=90-x \\ 2 x+x &=90 \\ 3 x &=90 \\ & x=90 / 3 \\ x &=30^{\circ} \end{aligned}$
$\begin{aligned} \text { Angles }=& x=30^{\circ} \\ & 90-x=\quad 60^{\circ} . \end{aligned}$
Question 10
Given
Adjacent angles ratio = 5:3
Let the adjacent angles = 5x, 3x
Sum of adjacent angles = 128$^{circ}$
$\therefore \begin{aligned} 5 x+3 x &=128^{\circ} \\ 8 x &=128 \\ x &=\frac{128}{8} \\ x &=16^{\circ} \end{aligned}$
∴ Adjacent angles = $5 x=5 \times 16=80^{\circ}$
$3 x=3 \times 16=48^{\circ}$
Question 11
(i) Sum of angles at a point = $360^{\circ}$
$\begin{aligned} x+41+105+130 &=360 \\ x+276 &=360 \\ x+276 &=360 \\ x &=360-276 \\ x &=84^{\circ} \end{aligned}$
(ii) Angles $3x^{\circ}, x^{\circ},40^{\circ}$ forms a linear pair
$\begin{aligned} 3 x+x+40^{\circ} &=180^{\circ} \\ 4 x+40 &=180 \\ 4 x &=180-40 \\ 4 x &=140 \\ x=& \frac{140}{4} \\ x &=35^{\circ} \end{aligned}$
(iii) Angles $2 x+10^{\circ}, \quad 3 x-10^{\circ}, 40^{\circ}$ forms a linear pair
$\begin{aligned} 2 x+10^{\circ}+3 x-10+40^{\circ} &=180 \\ 5 x+40 &=180 \\ 5 x &=180-40 \\ 5 x &=140 \\ x &=\frac{140}{5} \\ x &=28^{\circ} \end{aligned}$
Question 12
(i) $y=135^{\circ}$ (∵Vertically opposite angle)
Z = x (∵Vertically opposite angle)
$x, 135^{\circ}$ angles forms a linear pair
$\begin{aligned} x+135^{\circ} &=180^{\circ} \\ x &=180-135 \\ x &=45^{\circ} \end{aligned}$
$z=x=45^{\circ}$
$\therefore \quad x=45^{\circ}, \quad y=135^{\circ}, \quad z=45^{\circ}$
(ii) According to given diagram
$\begin{aligned} 31+y &=90^{\circ} \\ y &=90-31 \\ y &=59^{\circ} \end{aligned}$
x = $31^{\circ}$ (∵Vertically opposite angle)
$y=z=59^{\circ}(\because$ Vertically opposite angles $)$
$x=31^{\circ}, \quad y=59^{\circ}, z=59^{\circ}$
(iii) $x=44^{\circ}$ (∵Vertically opposite angle)
$z=51^{\circ}$ (∵Vertically opposite angle)
x, y , z Angles forms a linear pair
$x+y+z=180^{\circ}$
$44+y+51=180$
$y+95=180$
$y=180-95$
$y=85^{\circ}$
$\therefore x= 44^{\circ}, y=85^{\circ}, \quad z=51^{\circ}$
Question 13
From the given diagram
$\angle A F D=\angle C F B$ (∵ vertically opposite angles)
$\angle A F D=50^{\circ}$
Given $\quad \angle E F A=\angle A F D=50^{\circ}$
$\angle E F C+\angle A F D+\angle E F A=180^{\circ}(\because$ Forms a linear pair $)$
$\begin{aligned} \angle E F C+50+50 &=180 \\ \angle E F C+100 &=180 \\ \angle E F C &=180-100 \\ \angle E F C &=80^{\circ} \end{aligned}$
No comments:
Post a Comment