Exercise 11A
Question 1
In the figure ,drawn at the right AB = BC,AC =AE ,CE=AE. AF=EF$\angle A B C=90^{\circ}$
(IMAGE TO BE ADDED)
Name the triangle that are:
(i) Isosceles $\triangle A B C, \triangle A F E$
(ii) equilateral $\triangle A E C$
(iii) right angled $\triangle A B C, \triangle E D C$
(iv) right angled and isosceles $\triangle A B C$
Question 2
Find the unknown lettered angle.
(i) (IMAGE TO BE ADDED)
Sol:
By angle sum
Property
$\begin{aligned}&50+70+x=180^{\circ} \\&x=180^{\circ}-120^{\circ} \\&x=60^{\circ}\end{aligned}$
(ii) (IMAGE TO BE ADDED)
Sol: $\Rightarrow 65^{\circ}+65^{\circ}+\angle a=180^{\circ}$
$\Rightarrow \angle a=130^{\circ}-130^{\circ}$
$\Rightarrow \angle a=50^{\circ}$
(iii) (IMAGE TO BE ADDED)
Sol: $\Rightarrow x+x+x=180^{\circ}$
$\Rightarrow 3 x=180^{\circ}$
$\Rightarrow x=\frac{180^{\circ}}{3}$
$\Rightarrow x=60^{\circ}$ Answer
(iv) (IMAGE TO BE ADDED)
Sol: $50^{\circ}+70^{\circ}+y=180^{\circ}$
$\Rightarrow y=180^{\circ}-120^{\circ}$
$\Rightarrow y=60^{\circ}$
$x+y=180^{\circ}$
$\angle x=180^{\circ}-60^{\circ}$
$\angle x=120^{\circ}$
Question 3
In the following question, the measure of two angles are given . In each case find the measure of the third angle.
(i) $30^{\circ}, 80^{\circ}$
Sol: $30^{\circ}+80^{\circ}+x=180^{\circ}$
$x=180^{\circ}-110^{\circ}$
$x=70^{\circ} $ Answer
(ii) $40^{\circ}, 40^{\circ}$
Sol: $40^{\circ}+40^{\circ}+21=180^{\circ}$
$x=180^{\circ}-80^{\circ}$
$x=100^{\circ}$
(iii) $20^{\circ}, 70^{\circ}$
Sol: $20^{\circ}+70^{\circ}+21=180^{\circ}$
$x=180^{\circ}-90^{\circ}$
$x=90^{\circ} \mathrm{Ans}$
(iv) $59^{\circ}+45^{\circ}+21=180$
Sol:
$\begin{aligned} \Rightarrow x &=180^{\circ}-104^{\circ} \\ x &=76^{\circ} \text { Ans} \end{aligned}$
(v) $35^{\circ}+116^{\circ}+21=180^{\circ}$
$\Rightarrow x=180^{\circ}-151^{\circ}$
$\Rightarrow x=29^{\circ}$
Question 4
Find the measure of the angles of a triangle in each of the following cases:
(i)One of the acute angles of a right triangle is 63. find the other acute angle.
Ans: One angle = 90
So other acute angle + 63= 90
x$=90^{\circ}-63 \Rightarrow x=27^{\circ}$
(ii) The three angles are equal to one another
Ans: $x+x+x=180^{\circ} \Rightarrow 3 x=180^{\circ} \Rightarrow x=\frac{180^{\circ}}{3} \Rightarrow x=60^{\circ}$
(iii) One of the angles is $140^{\circ}$ and the other two angles are equal .
Sol: $x(x)+140^{\circ}=180^{\circ} \quad \Rightarrow \quad 2 x=180^{\circ}-140^{\circ}$
=$2 x=40^{\circ} \Rightarrow x=\frac{40^{\circ}}{2} \Rightarrow x=20^{\circ}$
(iv) One angle is twice the smallest angle and another angle is three times the smallest angle.
Sol: Let smallest angle = x ; $y=2 x ; \quad 2=3 x$
$x+y+z=180^{\circ} \quad ; \quad \Rightarrow \quad x+2 x+3 x=180^{\circ}$
$6 x=180^{\circ} \quad \Rightarrow \quad x=\frac{180^{\circ}}{6}$
$x=30^{\circ}$
$y=60$
z = $90^{\circ}$
(v) If one angle of a triangle is $80^{\circ}$ and other two angles are in the ratio 3: 7
Sol: $\Rightarrow 80^{\circ}+3 x+7 x=180^{\circ}$
$\Rightarrow 10 x=180^{\circ}-80^{\circ}$
$\Rightarrow 10 x=100$
$\Rightarrow \quad x=\frac{100}{10}=x=10^{\circ}$
$3 x=3 \times 10=30^{\circ}$
$7 x=7 \times 10=70^{\circ}$
(vi) The angle are in the ratio
2: 3: 4
2 x ; 3 x ; 4 x
Sol: $\Rightarrow 2 x+3 x+4 x=180^{\circ}$
$\Rightarrow \quad 9 x=180^{\circ}$
$\Rightarrow \quad x=\frac{180^{\circ}}{9} \Rightarrow x=20^{\circ}$
$2 x=20 \times 2=40^{\circ}$
$3 x=3 \times 20=60^{\circ}$
$4 x=4 \times 20=80^{\circ}$
Question 5
If each angle of a triangle is less than the sum of the other two ,show that the triangle is acute angled
Sol: Acc. to question $\angle A<\angle B+\angle C$
Add $\angle A$ both sides = $\angle A\angle A+<\angle A+\angle B+\angle C $
$2 \angle A<\angle A+\angle B+\angle C \Rightarrow 2 \angle A<180^{\circ}$
$\angle A<\frac{180^{\circ}}{2} \Rightarrow \angle A<90^{\circ}$Z Similarly
$\angle B<90^{\circ}$
$\angle C<90$
Question 6
In a $\triangle A B C$, if $3 \angle A=4 \angle B=6 \angle C$. Calculate the angles
Sol: Let $3 \angle A=4 \angle B=6 \angle C=x$
$\angle A=\frac{x}{3} ; \angle B=\frac{x}{4} ; \angle C=\frac{x}{6}$
$\angle A+\angle B+\angle C=180^{\circ} \Rightarrow \frac{x}{3}+\frac{x}{4}+\frac{x}{6} \equiv 180^{\circ}$
$\frac{8 x+6 x+4 x}{24}=180^{\circ} \Rightarrow 18 x=180^{\circ} \times 24$
$x=\frac{180^{\circ} \times 24}{18}$
$x=240^{\circ}$
$\angle A=\frac{x}{3}=\frac{240}{3}=80^{\circ} \quad ; \quad \angle B=\frac{x}{4}=\frac{240}{4}=60^{\circ}$
$\angle C=\frac{x}{6}=\frac{240}{6}=40^{\circ}$
Question 7
The given figures are drawn using more than one triangle. Find
(Image to be added)
Sol: (i) $\angle D A B+\angle A B C+\angle B C D+\angle C D A$
$\triangle A B C \quad \angle 1+\angle 3+\angle B=180^{\circ}$ ..................(1)
$\triangle A D C \Rightarrow \angle 2+\angle 4+\angle D=180^{\circ}$........(2)
$\Rightarrow \angle 1+\angle 2+\angle 3+\angle 4+\angle B+\angle D=180^{\circ}+180^{\circ}$
$\Rightarrow \angle D A B+\angle B C D+\angle A B C+\angle C D A=360^{\circ}$
(ii) $\angle F A B+\angle A B C+\angle B C D+\angle C D E+\angle D E F+\angle E F A$
Question 8
(IMAGE TO BE ADDED)
(i)In fig(i) Triangle ABC is right angled at A, and AD $\perp B C$ Also, $\angle B=55^{\circ}$. find
(a) $\angle B A D$
(b) $\angle C A D$
(c) $\angle A C B$
(a) $\triangle A B D: \rightarrow$
Sol: $\Rightarrow \angle B A D+\angle A B D+\angle A D B=180^{\circ} .$
$\Rightarrow \angle B A D+55^{\circ}+90^{\circ}=180^{\circ}$
$\Rightarrow \angle B A D=180^{\circ}-145^{\circ}$
$\Rightarrow \angle B A D=35^{\circ} \mathrm{Ans}$
(b)
$\begin{aligned} & \angle B A D+\angle C A D=90^{\circ} \\ \Rightarrow & 35+\angle C A D=90^{\circ} \Rightarrow \angle C A D=90-35^{\circ} \\ \Rightarrow & \angle C A D=55^{\circ} \text { Ans} \end{aligned}$
(c) $\angle A C B+\angle A D C+\angle D A C=180^{\circ}$
$\Rightarrow \quad \angle A C B+90^{\circ}+55^{\circ}=180^{\circ}$
$\Rightarrow \quad \angle A C B=180^{\circ}-145^{\circ} \Rightarrow \angle A C B=35^{\circ}$
(ii) Find the Lettered angles
(IMAGE TO BE ADDED)
Sol:
$\begin{aligned} \stackrel{\triangle A C D}{ } \rightarrow & \angle D A C+\angle A D C+\angle A C D=180^{\circ} \\ & 40^{\circ}+90^{\circ}+y=130^{\circ} \\ & \Rightarrow y=180^{\circ}-130^{\circ} \\ & \Rightarrow y=50^{\circ} \end{aligned}$
$\triangle A B D \rightarrow$
$\begin{aligned}&x+60^{\circ}+90^{\circ}=180^{\circ} \\&x=180^{\circ}-150^{\circ} \\&x=30^{\circ} \quad\end{aligned}$
Question 9
Is it possible to have a triangle in which
(i) Two of the angle are right angle ? No
(ii)Two of the angle are acute ? Yes
(iii)Two of the angles are obtuse? No
(iv) Each angle is less than $60^{\circ} ?$ No
(v)Each angle is greater than $45^{\circ}$ ? Yes
(vi) Each angle is equal to $60^{\circ} ?$ Yes
(vii) Each angle is greater than $60^{\circ} ?$ No
Question 10
(i) Find $\angle A M L$
(IMAGE TO BE ADDED)
Sol: $\angle A L M=\angle A B C=65^{\circ}$ [corresponding angle ]
Angle Sum $\angle A M L+35^{\circ}+65^{-0}=180^{\circ}$
$\angle A M L=180^{\circ}-100$
$\angle A M L =80^{\circ}$
(ii) Find $\angle T S Q$
(IMAGE TO BE ADDED)
Sol: $\angle P T S=\angle P R Q=80^{\circ}$
Angle sum = $\angle P S^{\circ} T^{\circ}+40+80^{\circ}=180^{\circ}$
$\angle P S T=180^{\circ}-120^{\circ}$
$\angle P S T=60^{\circ}$
By L. P $\angle{TSQ}=120^{\circ}$
(iii) find $\angle S L R$
(IMAGE TO BE ADDED)
Sol: $\Rightarrow$ by L.P. $\angle L S R=85^{\circ}$
$\Rightarrow \angle L R S=45^{\circ}$
Angle sum $\angle S L R + 45^{\circ}+85^{\circ}=180^{\circ}$
$\Rightarrow \angle S L R=180^{\circ}-130^{\circ}$
$\Rightarrow \angle S L R=50^{\circ}$
Question 11
Find (i) $\angle D B C$(ii) $\angle B D C$(iii) $\angle D B A$(iv) $\angle A$
(Diagram to be added)
(i) $\angle D B C=65^{\circ}$ [Alt] Ans
$\angle B D C+65^{\circ}+50^{\circ}=180^{\circ}$
$\Rightarrow \angle B D C=180^{\circ}-115$
(ii) $\angle B D C=65^{\circ}$
(iii) $\angle A B D=\angle B D C=65^{\circ} \quad$ [Alt] Ans
(iv) $\angle A+65^{\circ}+65^{\circ}=180^{\circ}$
$\angle A=180^{\circ}-130^{\circ}$
$\angle A=50^{\circ}$ Answer
Question 12
(i) $\angle B$ (ii) $\angle E D C$ (iii) $\angle A D E$(iv) $\angle A E D$(v) LDEC(vii) $\angle D C E$(vii) $\angle A C B$
(Diagram to be added)
(i)$\angle B+100+25=180$
$\angle B=180-125 \Rightarrow \angle B=55^{\circ}$ Answer
(ii)$\angle E D C=25^{\circ} \quad[\mathrm{All}]$
(iii) $\angle A D E+\angle E D C+\angle B D C=180^{\circ}$
$\angle A D E+25^{\circ}+100^{\circ}=180^{\circ}$
$\angle A D E=180^{\circ}-125^{\circ}$= $\angle A D E=55^{\circ}$
(iv) $\angle A E D+\angle A+\angle A D E=180^{\circ}$
Sol: $\Rightarrow \angle A E D+55^{\circ}+55^{\circ}=180^{\circ}$
$\Rightarrow \angle A E D=180^{\circ}-110^{\circ} \Rightarrow \angle A E D=70^{\circ}$
(v) $\angle D E C+25^{\circ}+45^{\circ}=180^{\circ} .$
$\angle D E C=180^{\circ}-70^{\circ}$
$\left.\angle D E C=110^{\circ}\right)$ Answer
(vi)
$\begin{aligned} & \angle A+\angle B+\angle C=180^{\circ} \\ \Rightarrow & 55^{\circ}+55^{\circ}+\angle C=180^{\circ}=\end{aligned}$ $\angle C=180^{\circ}-110^{\circ}$
$\Rightarrow \angle C=70^{\circ} \quad \Rightarrow \quad \angle B C D+\angle D C E=70^{\circ}$
$25^{\circ}+\angle D C E=70^{\circ}$
$\angle D C E=70^{\circ}-25^{\circ}$
$\angle D C E=45^{\circ}$ Ans
(vii) $\angle A C B=\angle C=70^{\circ}$ Answer
Question 13
Find (i) $\angle P T Q$ (ii) $\angle Q T U$ (iii) $\angle S U T$ (iv)$\angle RUQ$ (v) $\angle R Q U$ (vi) $\angle UQT$
(IMAGE TO BE ADDED)
(i) $\angle P T Q+90^{\circ}+20^{\circ}=180^{\circ}$
=$\angle P T Q=180^{\circ}-110^{\circ}$
$\Rightarrow \quad \angle P T Q=70^{\circ}$
(ii) $\angle P T Q+\angle Q T U+\angle S T U=180^{\circ}$
Sol: $\angle Q T U+45^{\circ}+70^{\circ}=180^{\circ}$
=$\angle Q T U=180^{\circ}-115^{\circ}$ =$\angle Q T U=65^{\circ}$ Answer
(iii) $\angle S U T+\angle T S U+\angle S T U=180^{\circ}$
Sol: $\Rightarrow \angle S U T+45^{\circ}+90^{\circ}=180^{\circ} \Rightarrow \angle S U T=180^{\circ}-135^{\circ}$
=$\angle S U T=45^{\circ}$ Answer
(iv) $\angle R U Q+\angle T U Q+\angle S U T=180^{\circ}$
Sol: $\Rightarrow \quad \angle RU Q+70^{\circ}+45^{\circ}=180^{\circ}$
$\angle R U Q=180^{\circ}-115^{\circ} \quad \Rightarrow \quad \angle RU Q=65^{\circ}$
(v)$\angle R Q U+\angle R U Q+\angle R=180^{\circ}$
Sol: $\angle R Q U+65^{\circ}+90^{\circ}=180^{\circ}$
$\angle R Q U=180^{\circ}-155^{\circ} \quad \Rightarrow \quad \angle R Q U=25^{\circ}$
(vi) $\angle U Q T+$ $\angle P Q T$ $+\angle R Q U=90$
Sol: $\Rightarrow \quad \angle U Q T+\angle P Q T+\angle R Q U=90^{\circ}$
$\Rightarrow \angle U Q T+20^{\circ}+25^{\circ}=90^{\circ}$
$\Rightarrow \quad \angle U Q T=90^{\circ}-45^{\circ} \Rightarrow \angle U Q T=45^{\circ}$ Answer
Question 14
Find $\angle P$
(DIAGRAM TO BE ADDED)
Sol: Given AB||PQ , AC||RP
Now draw line through R cutting AC ||BA
$\Rightarrow \angle C D R=\angle B A C=90^{\circ}$[corr. angle]
$\Rightarrow \angle C D R=\angle P R E=90^{\circ}$
$\Rightarrow \angle P R E=\angle R P Q=90^{\circ} \quad[$ Alt $]$ Answer
Question 15
In $\triangle A B C, \angle B=60^{\circ}, \angle C=40^{\circ}, A L \perp B C$ and AD bisects $\angle A$ such that $L$ and $D$ lie on side BC. find $\angle L A D$
(DIAGRAM TO BE ADDED)
Sol: $\angle B A L+\angle B+\angle B L A=180^{\circ}$
=$\angle B A L+60^{\circ}+90^{\circ}=180^{\circ}$
=$\angle B A L=180^{\circ}-150^{\circ}$
$\angle B A L=30^{\circ}$
$\angle C A L=50^{\circ}$
$\angle B A D=\angle D A C$
$\Rightarrow \quad 30+\angle \angle A D=50^{\circ}-\angle L A D$
$\Rightarrow \quad 2 \angle L A D=50-30^{\circ}$
$2 \angle LA D=20^{\circ}$
$\angle LA D=\frac{20}{2}$
$\angle LA D=10^{\circ}$
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