Exercise 11.2
Question 1
(i) Exterior angle =Sum of two interior opposite angles
$x=45+65$
$x=110^{\circ}$
(ii)Exterior angle = Sum two interior opposite angles
$x=55+40$
$x=95^{\circ}$
(iii) Exterior angle = Sum of two interior opposite angles
$x=50+50$
$x=100^{\circ}$
Question 2
(i)Exterior angle = Sum of two interior opposite angles
$115=x+50$
$x=115-50$
$x=65^{\circ}$
(ii) Exterior angle = Sum of two interior opposite angles
$80=x+30$
$x=80-30$
$x=50^{\circ}$
(iii) Exterior angle = Sum of two interior opposite angles
$70=x+36^{\circ}$
$x=70-36$
$x=34^{\circ}$
Question 3
(i) Exterior angle = Sum of two interior opposite angles
$105=x+2 x$
$3 x=105$
$x=\frac{105}{3}$
$x=35^{\circ}$
(ii) Exterior angle = Sum of two interior opposite angles
$\begin{aligned} 125 &=2 x+3 x \\ 125 &=5 x \\ x &=\frac{125}{5} \\ x &=25^{\circ} \end{aligned}$
(iii) Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} x+60+50 &=180^{\circ} \\ x+110 &=180 \\ x &=180-110 \\ x &=70^{\circ} \end{aligned}$
Question 4
(i) Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} 50+x+x &=180 \\ 2 x+50 &=180 \\ 2 x &=180-50 \\ 2 x &=130 \\ x &=\frac{130}{2} \end{aligned}$
x = $65^{\circ}$
(ii) Exterior angle = $110^{\circ}$, $120^{\circ}$
Corresponding interior angles = $180-110, \quad 180-120$
(diagram to be added)
= $70^{\circ},60^{\circ}$ (∵Forms linear pair)
∴ Angles of triangle x, $70^{\circ},60^{\circ}$
$\therefore \quad x+70+60=180$
$x+130=180$
$x=180-130 .$
$x=50^{\circ}$
(iii) Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} x+2 x+90 &=180 \\ 3 x+90 &=180 \\ 3 x &=180-90 \\ 3 x &=90 \\ x &=\frac{90}{3} \\ x &=30^{\circ} \end{aligned}$
Question 5
(i) Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} x+y+50 &=180 \\ x+y &=180-50 \\ x+y &=130 \rightarrow(①\end{aligned}$
Exterior angles= Sum of interior angles
$120=50+x$
$x=120-50$
$x=70^{\circ} \rightarrow$②
Substituted x value in eq ①
$x+y=130$
$70+y=130$
$y=130-70$
$y=60^{\circ}$
(ii) x $=60^{\circ}$ (∵Vertically opposite angles)
Sum of angles in triangles $=180^{\circ}$
$40+x+y=180^{\circ}$
$40+60+y=180^{\circ}$
$y+100=180$
$y=180-100$
$y=80^{\circ}$
(iii) $y=90^{\circ}$ (∵Vertically opposite angles)
Sum of angles in triangles $=180^{\circ}$
$\begin{aligned} y+x+x &=180^{\circ} \\ 90+2 x &=180^{\circ} \\ 2 x &=180-90 \\ 2 x &=90 \\ x &=\frac{90}{2} \\ x &=45^{\circ} \end{aligned}$
Question 6
(i) (diagram to be added)
Angles of triangle = x , x ,y (∵Vertically opposite angles are equal)
x= y (∵Vertically opposite angles are equal)
Sum of angles in triangle $=180^{\circ}$
$\begin{aligned} x+x+y &=180 \\ x+x+x &=180 \\ 3 x &=180 \\ x &=\frac{180}{3} \\ x &=60^{\circ} \end{aligned}$
(ii) From given figure
$125+x=180^{\circ}$ (∵Forms linear pair)
$x=180-125$
$x=55^{\circ}$
Sum of angles in triangle $=180^{\circ}$
$\begin{aligned} x+x+y &=180^{\circ} \\ 2 x+y &=180 \\ 2 \times 55+y &=180 \\ 110+y &=180 \\ y &=180-110 \\ y &=70^{\circ} \end{aligned}$
(iii) (diagram to be added)
In triangle ADC
$\begin{aligned} 50+x+70=& 180^{\circ} \\ x+120 &=180 \\ x &=180-120 \\ x &=60^{\circ} \end{aligned}$
In triangle ABC
$\begin{aligned}(50+y)+36+70 &=180 \\ y+156 &=180 \\ y &=180-156 \\ y &=24^{\circ} \end{aligned}$
Question 7
(diagram to be added)
Sum of angles in triangle $=180^{\circ}$
$\begin{aligned} x+80+52 &=180 \\ x+132 &=180 \\ x &=180-132 \\ x &=48^{\circ} \end{aligned}$
z + 143 = $180^{\circ}$ (∵Forms a linear pair)
$x=180-143$
$x=37^{\circ}$
Sum of angles in triangle = $180^{\circ}$
$x+y+z=180$
$48^{\circ}+37^{\circ}+z=180$
$85+z=180$
$z=180-85$
$z=95^{\circ}$
Question 8
Let the unknown angles are x,x
One angle = $80^{\circ}$
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} x+x+80=& 180 \\ 2 x+80 &=180 \\ 2 x &=180-80 \\ & 2 x=100 \\ & x= \frac{100}{2} \\ & x=50^{\circ} \end{aligned}$
Question 9
Given
Angle in triangle = $60^{\circ}$
Other two angles ratio = 2: 3
Let two angles = 2x, 3x
Sum of angles = 180
$60+2 x+3 x=180$
$\begin{aligned} 60+5 x &=180 \\ 5 x &=180-60 \\ 5 x &=120 \\ x=& \frac{120}{5} \\ x &=24^{\circ} \end{aligned}$
$\begin{aligned} \therefore \text { Angles are } & 2 x=2 \times 24=48^{\circ} \\ 3 x &=3 \times 24=72^{\circ} \end{aligned}$
Question 10
Given
Angles ratio = 1: 2: 3
Let angles be x, 2x , 3x
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} x+2 x+3 &=180^{\circ} \\ 6 x &=180 \\ x &=\frac{180}{6} \\ x &=30^{\circ} \end{aligned}$
Angle = x = $30^{\circ}$
$2 x=2 \times 36^{\circ}=66^{\circ}$
$3 x=3 \times 30^{\circ}=90^{\circ}$
∴ It is scalene triangle according to sides
It is right angled triangle according to angles
Question 11
(i)$65^{\circ}, 74^{\circ}, 39^{\circ} ?$
Sum of angles = $65^{\circ}+74^{\circ}+39=178^{\circ} \neq 180^{\circ}$
No, these are not angles of triangle
(ii)$\frac{1}{3}$ right angle $=\frac{1}{3} \times 90=30^{\circ}$
Right angle = $90^{\circ}$
Another angle= $60^{\circ}$
Sum of angles = $30+90+60=180^{\circ}$
∴ These are angles of triangle
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