Exercise 11B
Question 1
Find the lettered angle in each case:
(i)(IMAGE TO BE ADDED)
Sol: $\Rightarrow x+30^{\circ}=70^{\circ}$
$\Rightarrow x=70-30^{\circ}$
$\Rightarrow x=40^{\circ}$
(ii) (IMAGE TO BE ADDED)
Sol:
$\begin{aligned} \angle a &=35^{\circ}+50^{\circ} \\ \angle a &=85^{\circ} \end{aligned}$
(iii) (IMAGE TO BE ADDED)
Sol: $\angle y=30+45$
$\angle y=75^{\circ}$
(iv) (IMAGE TO BE ADDED)
Sol: $\Rightarrow 100^{\circ}=x+70^{\circ}$
$\Rightarrow x=100-70$
$\Rightarrow x=30^{\circ}$
Question 2
Find the lettered angle in each case:
(IMAGE TO BE ADDED)
(i) By l.p y+ $120^{\circ}=180^{\circ}$
$y=180^{\circ}-120^{\circ}$
$y=60^{\circ}$ Ans
$x=40+60 \Rightarrow y=100^{\circ}$
(ii) (IMAGE TO BE ADDED)
Sol: $\triangle A P B \quad \angle P+\angle P A B+\angle P B A=180^{\circ}$
$\Rightarrow \quad \angle P B A+50^{\circ}+100^{\circ}=130^{\circ}$
$\Rightarrow \angle P B A=130^{\circ}-150^{\circ}$
$\Rightarrow \angle P B A=30^{\circ}$ Ans
$\Rightarrow \dot{y}=50^{\circ}+2 \Rightarrow y=50+30$
$\Rightarrow y=80^{\circ}$ Ans
$\Rightarrow x+y+75^{\circ}=180^{\circ} \Rightarrow x+80^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow x=180^{\circ}-155^{\circ} \Rightarrow x=25^{\circ} \Rightarrow \mathrm{Ans}$
(iii) (IMAGE TO BE ADDED)
$x=70^{\circ}$ [Vertically]
$x+y=110^{\circ}$ $\Rightarrow y=110^{\circ}-70^{\circ} \Rightarrow y=40^{\circ}$
$\Rightarrow \quad x+y+z=180^{\circ}$
$\Rightarrow 70^{\circ}+40^{\circ}+2=180^{\circ} \Rightarrow z=180^{\circ}-110^{\circ} \Rightarrow 2=70^{\circ}$
(iv) (IMAGE TO BE ADDED)
$90^{\circ}=50^{\circ}+y$
$\Rightarrow y=90^{\circ}-50^{\circ} \Rightarrow y=40^{\circ}$
$\triangle D E C$
$\begin{aligned} 100+40 &+x=180^{\circ} \\ x &=180-140 \\ x &=40^{\circ} \mathrm{Ans} \end{aligned}$
Question 3
One of the exterior angles of a triangle is 100 degree and its interior opposite angles are equal to each other . what is the measure of each of these two angles?
Sol: Let angles =x
$\Rightarrow x+x=100 \Rightarrow 2 x=100 \Rightarrow x=50^{\circ}$
Question 4
One of the exterior angle of a triangle is $130^{\circ}$ and the interior opposite angles are in the ratio 2: 3 . find the angles of the triangle .
Sol: $2 x+3 x=130^{\circ} \quad \Rightarrow 5 x=130^{\circ}$
$\Rightarrow x=\frac{130}{5} \quad \Rightarrow \quad x=26^{\circ}$
$\Rightarrow 2 x=2 \times 26=52^{\circ}$ Ans $; \quad 3 x=3 \times 26=78^{\circ}$ Ans
Question 5
Find the value of x in each case:
(IMAGE TO BE ADDED)
Sol: $\triangle A B C$
$60+45^{\circ}+\angle C=180^{\circ}$
$\angle C=180^{\circ}-105^{\circ}$
$\angle C=75^{\circ}$ Ans
$\triangle C E D$
$\begin{aligned} & 80+75+x=180^{\circ} \\ \Rightarrow & x=180^{\circ}-155^{\circ} \Rightarrow x=25^{\circ} \end{aligned}$
(ii) (IMAGE TO BE ADDED)
Sol: Ex= Int opp + int opp
$\Rightarrow 11 x-60^{\circ}=3 x+40^{\circ}+4 x$
$\Rightarrow 11 x-60^{\circ}=7 x+40^{\circ}$
$\Rightarrow 11 x-7 x=60^{\circ}+40^{\circ}$
$4 x=100^{\circ}=x \cdot x=\frac{100}{4}$
$x=25^{\circ}$ Ans
(iii) (IMAGE TO BE ADDED)
By L.P
$\Rightarrow 100+y=180^{\circ}$
$\Rightarrow y=180^{\circ}-100^{\circ}$
$\Rightarrow y=80^{\circ}$ Ans
$\Rightarrow y+3 x^{\circ}=7 x^{\circ}$
$\Rightarrow 30^{\circ}=7 x-3 x \Rightarrow 4 x=30^{\circ}$
x = $20^{\circ}$ Answer
(iv) (IMAGE TO BE ADDED
Sol: $\angle P T Q=\angle T Q R+\angle Q R T$
$90^{\circ}=48^{\circ}+\angle Q R T$
$\Rightarrow \quad \angle Q R T=90-48^{\circ}$
$\Rightarrow \angle Q R T=42^{\circ}$
$\triangle P S R \rightarrow$
$x=\angle S P R+\angle S R P$
$x=30^{\circ}+42^{\circ}$
$x=72^{\circ}$ Ans
No comments:
Post a Comment