Exercise 11.3
Question 1
(i) (diagram to be added)
In, triangle angles opposite to equal sides are equal
∴ x = $50^{\circ}$
(ii) (diagram to be added)
Exterior angle + interior angle $=180^{\circ}$
$\begin{aligned} 110^{\circ}+& \angle B A C=180 \\ \angle B A C &=180-110 \\ & \angle B A C=70^{\circ} \end{aligned}$
In triangle , angles opposite to equal sides are equal
$\therefore \angle A B C=\angle A C B=x$
Sum of angles in triangle = $180^{\circ}$
$\angle A B C+\angle A C B+\angle B A C=180^{\circ}$
$x+x+70=180^{\circ}$
$2 x=180-70$
$2 x=110$
$x=\frac{116}{2}$
$x=55^{\circ}$
(iii)In triangle , angles opposite to equal sides are equal
$\angle B A C=\angle A C B=x .$
x =$36^{\circ}$ (∵Vertically opposite angles)
Question 2
(i) (diagram to be added)
In a triangle , angles opposite to equal sides are equal
$\angle B A C=\angle A C B=x$
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} 40+x+x &=180 \\ 40+2 x &=180 \\ 2 x &=180-40 \\2 x &=140 \\ x &=\frac{140}{2} \end{aligned}$
x = $70^{\circ}$
(ii) (diagram to be added)
In a triangle, angles opposite to equal sides are equal
$\angle B C A=\angle C A B$
$\angle C A B=45^{\circ}$
Sum of angles in triangle = $180^{\circ}$
$\angle B C A+\angle C A B+\angle A B C=180^{\circ}$
$\begin{aligned} 45+45+\angle A B C=180 \\ 90+\angle A B C &=180 \\ \angle A B C &=180-90 \\ \angle A B C &=90^{\circ} \end{aligned}$
(iii) (diagram to be added)
$l \| B C$, AB is transversal line
$\angle C B A=\angle P A Q$
$\angle C B A=50^{\circ}$
In a triangle , angles opposite to equal sides are equal
$\angle B C A=\angle B A C=x$
Sum of angles in triangle = $180^{\circ}$
$\angle B C A+\angle B A C+\angle C B A=180^{\circ}$
$x+x+50=180$
$2 x+50=180$
$2 x=180-50$
$2 x=130$
$x=\frac{130}{2}$
$x=65^{\circ}$
Question 3
(i)(diagram to be added)
In a triangle
Sum of exterior and interior angle = $180^{\circ}$
$\angle A C B+\angle A C D=180^{\circ}$
$\angle A C B+120=180$
$\angle A C B=180-120$
$\angle A C B=60^{\circ}$
In a triangle, angles opposite to equal sides are equal
$\angle A B C=\angle A C B .$
$y=60^{\circ}$
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} \angle A B C+\angle B C A+ \angle C A B=180^{\circ} \\ 60+60+ x=180^{\circ} \\ x+120 &=180 \\ x &=180-120 \\ \quad x &=60^{\circ} \end{aligned}$
(ii) (diagram to be added)
In a triangle
Exterior angle + Interior angle = $180^{\circ}$
$\begin{aligned} 115^{\circ}+\angle A C B &=180^{\circ} \\ \angle A C B &=180-115 \\ \angle A C B &=65^{\circ} \end{aligned}$
In a triangle,
Angles opposite to equal sides are equal
$\angle A C B=\angle B A C=65^{\circ}$
$x=65^{\circ}$
Sum of angle in a triangle = $180^{\circ}$
$\angle A B C+\angle A C B+\angle B A C=180$
$y+65^{\circ}+65^{\circ}=180$
$y+130=180$
$y=180-130$
$y=50^{\circ}$
(iii) (diagram to be added)
$\angle A B D=\angle A D B=x$ (∵In a triangle, angles opposite to the equal sides are equal)
Sum of angle in triangle $=180^{\circ}$
$\begin{aligned} x+x+48=& 180 \\ 2 x+48 &=180 \\ 2 x &=180-48 \\ 2 x &=132 \\ x &=\frac{132}{2} \\ x &=66^{\circ} \end{aligned}$
External angle + Internal angle = $180^{\circ}$
$\angle A D C+66=180^{\circ}$
$\angle A D C =180-66$
$\angle A D C=114^{\circ}$
In triangle ADC
$\angle A C D=\angle D A C=y$ (∵In a triangle, angles opposite to equal sides are equal)
Sum of angles in triangle = 180
$\begin{aligned} y+y+114^{\circ} &=180^{\circ} \\ 2 y+11 y^{\circ} &=180^{\circ} \\ 2 y &=180-114 \\ 2 y &=66 \\ y &=\frac{66}{2} \\ y &=33^{\circ} \end{aligned}$
Question 4
(diagram to be added)
In triangle ADB
$\begin{aligned} \angle B A D &=\angle A B D(\because \overline{A D}=\overline{B D}) \\ x &=32^{\circ} \end{aligned}$
In triangle BDC
$\angle B C D=\angle B D C=y \quad(\because \overline{B D}=\overline{B C})$
Sum of angles in triangle =$180^{\circ}$
$\begin{aligned} \angle B D C+\angle D B C &+\angle D C B=180^{\circ} \\ y+z+y &=180^{\prime} \\ z+2 y &=180\end{aligned}$ -------①
In triangle ABC
$\begin{aligned} \angle A+\angle A B C+\angle A B C A &=180^{\circ} \\ 32+x+z+y &=180^{\circ} \\ 32+32+z+y &=180^{\circ} \\ z+y &=180-64 \end{aligned}$
$z+y=116 \rightarrow(2)$
Solving 1 and 2
(diagram to be added)
$z+y=116$
$z+64=116$
$z=116-64$
$z=52^{\circ}$
(ii) (diagram to be added)
In triangle ABD
$\angle A B D=\angle B A D=(∵\overline{B D}=\overlineA D)$
$\angle B A D=35^{\circ}$
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} \angle A B D+\angle B A D+\angle A D B &=180 \\ 35+35+\angle A D B &=180 \\ \angle A D B &=180-70 \\ \angle A D B &=110^{\circ} \end{aligned}$
External angle + Internal angle = 180
$\begin{aligned} 110+\quad y &=180 \\ y &=180-110 \\ y &=70^{\circ} \end{aligned}$
In triangle ADC
$\begin{aligned} \angle A D C &=\angle A C D \\ y &=\angle A C D \end{aligned}$
External angle = Sum of other two interior angle
$x=\angle A D C+\angle A C D$
$x=y+y$
$x=2 y$
$x=2 \times 70$
$x=140^{\circ}$
Question 5
Given
Angles ratio = 1:2:1
;Let Angles $=x, 2 x, x$
Sum of angles = 180
$\begin{aligned} x+2 x+x &=180 \\ 4 x &=180 \\ x &=\frac{180}{4} \\ x &=45^{\circ} \end{aligned}$
∴ Angles= $45^{]circ}, 2 \times 45^{\circ}, 45$
$=45^{\circ}, 90^{\circ}, 45^{\circ}$
Question 6
(diagram to be added)
Both base angles are equal because Isosceles triangle
x -- base angle
y -- Vertical angle
Given
Base angle = $4 \times$ vertical angle
x=4 y
Sum of angles in triangle = $180^{\circ}$
$\begin{aligned} y+x+x=& 180 \\ y+4 y+4 y &=180^{\circ} \\ 9 y &=180 \\ y &=180 / 9 \\ y &=20^{\circ} \end{aligned}$
Angles = y,4y, 4y = $20^{\circ}, 80^{\circ}, 80^{\circ}$
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