S Chand Class 10 CHAPTER 14 Circle Exercise 14 F

 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