Exercise 14 F
Question 1
Ans: In the figure, $P Q$ is the tangent to the circle at $\angle B T P=75^{\circ}$ and $\angle A T B=45^{\circ}$
But $\angle Q T A+\angle A T B+\angle B T P=180^{\circ}$
$\angle Q T A+45^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow \angle Q T A+120^{\circ}=180^{\circ}$
$\Rightarrow \angle Q T A=180^{\circ}-120^{\circ}=60^{\circ}$
Now $\angle Q T A$ is the tangent and $T A$ is the chord
So $\angle Q T A=\angle A B T$
$\Rightarrow \angle A B T=60^{\circ}$
Question 2
Ans: In the figure a circle with center 0 TAS is a tangent to the circle at $A, A B, A C$ and $B C$ are chords the circle $\angle O B A=32^{\circ}$
In $\triangle O A B$
$O A=O B$
So $\angle O A B=\angle O B A=32^{\circ}$
In $\triangle A O B$ $\angle A O B+\angle O A B+\angle O B A=180^{\circ}$ $\Rightarrow \angle A O B+32^{\circ}+32^{\circ}=180^{\circ}$ $\Rightarrow \angle A O B+64^{\circ}=189$ $\Rightarrow \angle A O B=180^{\circ}-64=116^{\circ}$
$\begin{aligned}&\text { In } \triangle A O B \\&\angle A O B+\angle O A B+\angle O B A=180^{\circ} \\&\Rightarrow \angle A O B+32^{\circ}+32^{\circ}=180^{\circ} \\&\Rightarrow \angle A O B+64^{\circ}=188 \\&\Rightarrow \angle A O B=180^{\circ}-6+=116^{\circ}\end{aligned}$
Now Arc AB subtends $\angle A O B$ at the center and $\angle A C B$ at the remaining part of the circle So $\angle A O B=2 \angle A C B$
$\Rightarrow 116^{\circ}=2 y \Rightarrow y=\frac{116^{\circ}}{2}=58^{\circ}$
But $S T$ is the fangent and $A B$ is the chord
So $\angle B A S=\angle A C B$
$\Rightarrow x=y \Rightarrow x=y=58^{\circ}$
Question 3
Ans: In the circle with center O ,
$L N$ is the diameter
$P Q$ il a tangent to the circle at K
$\angle K L N=30^{\circ}$ and $\angle M N L=60^{\circ}$
In $\triangle L K N$
$\angle L K N=90^{\circ}$
$\angle K L N=30^{\circ}$
So $\angle K N L=90^{\circ}-30^{\circ}=60^{\circ}$
(i) Now PQ is tangent and KN is chord of the circle at K
So $\angle Q K N=\angle K L N=30^{\circ}$
(ii) Again $P Q$ is tangent and $K L$ is the chord
$\text { So } \angle P K L=\angle K N L=60^{\circ}$
(iii) In Cyclic quad ∠ MNK
$\angle K N M+\angle K L M=180^{\circ}$
$\Rightarrow \angle K N L+\angle L N M+\angle M L K=180^{\circ}$
$\Rightarrow 60^{\circ}+60^{\circ}+\angle M L K=180^{\circ}$
$\Rightarrow \angle M L K+120^{\circ}=180^{\circ}$
$\Rightarrow \angle M L K=180^{\circ}-120^{\circ} \Rightarrow \angle M L K=60^{\circ}$
Question 3
Ans: In the figure AT is the tangent to the circle and $\angle A B C=50^{\circ}$
$A C=B C$
So $\angle B A C=\angle A B C=50^{\circ}$
But $\angle A C B+\angle B A C+\angle A B C=180^{\circ}$
$\angle A C B+50^{\circ}+50^{\circ}=180^{\circ}$
$\Rightarrow \angle A C B 7100^{\circ}=180^{\circ}$
$\Rightarrow \angle A C B=180^{\circ}-100^{\circ}=80^{\circ}$
if $A T$ is the tangent and $A B$ is chord
so $\quad \angle B A T=\angle A C B=80^{\circ}$
Question 5
Ans: In the figure O is the center of circumcircle of $\triangle XYZ$
Tangents at X and Y are drawn to meet at T $\angle X T Y$
$=80^{\circ} \text { and } \angle \times 02=140^{\circ}$
Join OY
$X T$ and $Y T$ are tangents to the circle
So $\quad \angle \times T y+\angle \times 04=180^{\circ}$
(IMAGE TO BE ADDED)
$\Rightarrow 80^{\circ}+\angle XOY=180^{\circ}$
$\Rightarrow \quad \angle XOY \Rightarrow 180^{\circ}-80^{\circ}=100^{\circ}$
But $\angle XOY+\angle YOZ +\angle ZOX=360^{\circ}$
$\Rightarrow 100^{\circ}+\angle YOZ+140^{\circ}=360^{\circ}$
$\Rightarrow \quad \angle YOZ +240^{\circ}=360^{\circ}$
So $\angle YOZ=360^{\circ}-240^{\circ}=120^{\circ}$
Now are YZ subtends $\angle YOZ$ at the center and $\angle ZXY$
at the remaining part of the circle
So $\angle YOZ=2 \angle ZXY$
$\Rightarrow \quad \angle Z \times y=\frac{1}{2} \angle 402=\frac{1}{2} \times 120^{\circ}=60^{\circ}$
Question 6
Ans: O is the center of the circle AB is the chord and QBS is the tangent to the circle at B $\angle A B=110^{\circ}$
Arc AB subtends $\angle A O B$ at the center and $\angle A P B$ at the remaining part of the circle
So $\angle A O B=2 \angle A P B \Rightarrow \angle A P B=\frac{1}{2} \quad 2 A O B$
$\Rightarrow \angle A P B=\frac{1}{2} \times 110^{\circ}=55^{\circ}$
Now $Q B S$ is the tangent and $A B$ is the chord
So $\angle A B Q=\angle A P B=55^{\circ}$
Hence $\angle A P B=\angle A B Q=55^{\circ}$
Question 7
Ans: In a circle with center O and AB is a AT C, a tangent is drawn to the circle which meet AB on producing at D
$\angle B A C=30^{\circ}$
To prove: $B C=B D$
Construction: Join $B C$
proof: $C D$ is tangent and $C B$ is chord
So $\angle D C B=\angle B A C=30^{\circ}$
In $\triangle A B C$
$\begin{aligned}&\angle A C B=90^{\circ} \\&\text { so } \angle B A C+\angle C B A=90^{\circ} \\&\Rightarrow 30^{\circ}+\angle C B A=90^{\circ} \\&\Rightarrow \angle C B A=90^{\circ}-30^{\circ}=60^{\circ}\end{aligned}$
Now in $\triangle B C D$
Ext. $\angle C B A=\angle B C D=\angle B C D+\angle B D C$
$\Rightarrow 60^{\circ}=30^{\circ}+\angle B D C \Rightarrow \angle B D C=60^{\circ}-30^{\circ}=30^{\circ}$
if $\angle B C D=\angle B D C=30^{\circ}$
so $B C=B D \quad$
Question 8
Ans: In a circle DE is a tangent at A to the circumcircle of $\triangle A B C$ in which $D E \| B C$
To prove: AB= AC
Proof: IF DE is the tangent and AB is the chord of the circle
So $\angle D A B=\angle A C B$ ...........(i)
But $D E \| B C$
So $\angle D A B=\angle A B C$
from ii) and (ii)
$\begin{aligned}&\angle A C B=\angle A B C \\&\text { So } A B=A C\end{aligned}$ Hence proved
Question 9
Ans: Two circle intersect each other at B and C. Lines ABD and ACE are drawn to meet the smaller circle at D and E respectively . AF is the tangent to the first circle at A
To prove: AF || DE
construction: Join BC
proof : $A F$ is the tangent and $A B$ is the chord of the first circle
So $\angle B A F=\angle A C B$
In cyclic quadrilateral BCED.
$E \times t \cdot \angle A C D=$ int $O P P \cdot \angle B D E$.......(ii)
from (i) and (ii)
$\angle B A F=\angle B O E$
But these are alternate angles
So AF $\| D E$
Hence Proved
Question 10
Ans: A circumcircle of $\triangle A B C$ a tangent $D C$ is drawn at and $A B$ is Produced to meet the tangent at $D$
To prove: $\triangle D B C-\triangle D C A$
Proof: CD is tangent and BC is the chord of the circle
So$\angle B C D=\angle B A C$
Now in $\triangle D B C$ and $\triangle D C A$.
$\angle D=\angle D$
$\angle B C D=\angle B A C$ or $\angle D A C$
So $\triangle A B C \sim \triangle D C A$ Hence proved
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