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#### To draw a pair of tangents to a circle which are inclined to each other at an Angle of $60^{\circ}$, it is required to draw tangents at end points of those two radii of The circle, the angle between them should be(A) $135^{\circ}$                      (B) $90^{\circ}$                        (C) $60^{\circ}$                      (D) $120^{\circ}$

Solution
According to question:-

Given :$\angle QPR= 60^{\circ}$
Let       $\angle QOR= x$
As we know that angle between tangent and radius of a circle is 90

$\angle PQO= \angle PRO= 90^{\circ}$
We know that $\angle PQO+ \angle PRO+ \angle QPR+\angle QOR= 360^{\circ}$

[$\because$ sum of interior angles of quadrilateral is $360^{\circ}$ ]

$90^{\circ}+90^{\circ}+x+60^{\circ}= 360^{\circ}$|
$240+x= 360^{\circ}$
$x= 120^{\circ}$

#### To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{8}{5}$ of the corresponding sides of $\bigtriangleup ABC$ draw a ray BX such that $\angle CBX$ is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is (A) 5                            (B) 8                            (C) 13                          (D) 3

Solution
To construct a triangle similar to a triangle, with its sides  $\frac{8}{5}$  of the corresponding sides of given triangle, the minimum number of points to be located at an equal distance is equal to the greater of 8 and 5 in  .$\frac{8}{5}$ Here $8> 5$
So, the minimum number of points to be located at equal distance on ray BX is 8.

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#### To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides of $\frac{3}{7}$ the corresponding sides of $\bigtriangleup ABC$, first draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and next step is to join(A) B10 to C                (B) B3 to C                  (C) B7 to C                  (D) B4 to C

Solution Given: $\angle CBX$ is an acute angle.

Steps of construction
1.   Draw any ray BX making angle with BC on the side opposite to vertex A.
2.   Locate 7 points on BX in equidistant
3.   Now join Bto C

4.   Draw a line through B3${C}'$  parallel to B7C .

#### To divide a line segment AB in the ratio $5:6$, draw a ray AX such that $\angle BAX$ is an acute angle, then draw a ray BY parallel to AX and the points $A_{1},A_{2},A_{3}\cdots$and $B_{1},B_{2},B_{3}\cdots$are located at equal distances on ray AX and BY, respectively. Then the points joined are (A) A5 and B6            (B) A­ and B5               (C) A4 and B5             (D) A5 and B4

Solution

Given: $\angle BAX$ and $\angle ABY$ both are acute angles and AX parallel to BY
The required ratio is $5:6$
Let m = 5 , n = 6

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray $BY\parallel AX$.
3. Locate the points $A_{1},A_{2},A_{3},A_{4},A_{5}$ on AX at equal distances
4. Locate the points $B_{1},B_{2},B_{3},B_{4},B_{5}$ on BY at distance equal to the distance between points on AX line.
5. Join $A_{5}B_{6}$.
Let it intersect AB at a point C in figure.
Then $AC:CB= 5:6$
Here $\bigtriangleup AA_{5}C$  is similar to $\bigtriangleup BB_{6}C$
Then $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}= \frac{AC}{BC}$
$\therefore$by construction $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}$
$\therefore \frac{AC}{BC}= \frac{5}{6}$
$\therefore$ Points joined one A5 and B6.

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#### To divide a line segment AB in the ratio $4:7$, a ray AX is drawn first such that $\angle BAX$ is an acute angle and then points $A_{1},A_{2},A_{3}\cdots$ are located at equal distances on the ray AX and the point B is joined to:(A) A12                        (B) A11            (C) A10            (D) A9

Solution

Given: $\angle BAX$ is an acute angle
The required ratio is $4:7$
Let m = 4, n = 7
$m+n= 4+7= 11$

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 11 points on AX at equal distances   (because m + n = 11)
3. Join $A_{11}B$
4. Through the point A4 draw a line parallel to $A_{11}B$  intersecting AB at the point P.
Then $AP:PB= 4:7$

Hence point B is joined to A11.

#### To divide a line segment AB in the ratio $5:7$, first a ray AX is drawn so that$\angle BAX$ is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is (A) 8                            (B) 10                          (C) 11                          (D) 12

Solution
Given: $\angle BAX$ is an acute angle.
The required ratio is $5:7$
Let m = 5, n = 7

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances.   (Because here $m+n= 12$)
3. Join $A_{12}B$
4. Through the point $A_{5}$ draw a line parallel to  $A_{12}B$ intersecting AB at the point P.
Then $AP:PB= 5:7$
$\because A_{5}P\parallel A_{12}B$
$\therefore \frac{AA_{5}}{A_{5}A_{12}}= \frac{AP}{PB}$  (By Basic Proportionality theorem)

By construction $\frac{AA_{5}}{A_{5}A_{12}}= \frac{5}{7}$
$\therefore \frac{AP}{PB}= \frac{5}{7}$
Hence the number of points is 12.

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#### In Fig. 10.9, $\angle$AOB = 90º and $\angle$ABC = 30º, then $\angle$CAO is equal to:(A) 30º(B) 45º(C) 90º(D) 60º

(D)

Solution:

In AOB,

$\angle$OAB +$\angle$ABO + $\angle$BOA = 180°      … (i)                           (angle sum property of Triangle)

Angles opposite to equal sides are equal

$\angle$OAB = $\angle$ABO

Equation (i) becomes

$\angle$OAB + $\angle$OAB + 90° = 180°

2$\angle$OAB = 180° – 90°

$\angle$OAB  = 45° …(ii)

In $\triangle$ACB,

$\angle$ACB + $\angle$CBA + $\angle$CAB = 180°   (angle sum property of Triangle)

$\angle ACB = \frac{1}{2} \angle BOA = 45^{\circ}$

(Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)

$\therefore$  45° + 30° + $\angle$CAB = 180°

$\angle$CAB = 180° – 75° = 105°

$\angle$CAO + $\angle$OAB = 105°

$\angle$CAO + 45° = 105°

$\angle$CAO = 105° – 45° = 60°

Therefore option (D) is correct.

#### In Fig. 10.8, BC is a diameter of the circle and $\angle$BAO = 60º. Then $\angle$ADC is equal to:Fig. 10.8(A) 30º(B) 45º(C) 60º(D) 120º

(C)

Solution:

In $\triangle$AOB, AO = OB    (Radius)

$\angle$ABO = $\angle$BAO         [Angle opposite to equal sides are equal]

$\angle$ABO = 60° [$\therefore$ $\angle$BAO = 60° Given]

In $\triangle$AOB,

$\angle$ABO + $\angle$OAB =$\angle$AOC (exterior angle is equal to sum of interior opposite angles)

60° + 60° = 120° =$\angle$AOC

$\angle ADC =\frac{1}{2} \angle AOC$

(Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)

$\angle ADC =\frac{1}{2} \left ( 120^{\circ} \right )$

$\angle$ADC = 60°

Therefore option (C) is correct.

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#### ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle$ADC = 140º, then $\angle$BAC is equal to:(A) 80º(B) 50º(C) 40º(D) 30º

(B) 50°

Solution:

Given, ABCD is cyclic Quadrilateral and $\angle$ADC = 140°

We know that the sum the opposite angles in a cyclic quadrilateral is 180°.

$\angle$ADC + $\angle$ABC = 180°

140° + $\angle$ABC = 180°

$\angle$ABC = 180° – 140°

$\angle$ABC = 40°

Since, $\angle$ACB is an angle in semi circle

$\angle$ACB = 90°

In $\triangle$ABC

$\angle$BAC + $\angle$ACB + $\angle$ABC = 180° (angle sum property of a triangle)

$\angle$BAC + 90° + 40° = 180°

$\angle$BAC = 180° – 130° = 50°

Therefore option (B) is correct.

#### In Fig. 10.7, if $\angle DAB = 60 ^{\circ}, \angle ABD = 50^{\circ}$, then $\angle ACB$ is equal to:Fig. 10.7(A) 60º(B) 50º(C) 70º(D) 80º

(C) 70°

Solution:

In $\triangle ABD$,

$\angle ABD + \angle ADB +\angle DAB = 180^{\circ}$

(Angle sum property of triangle)

$\\50^o+ < ADB + 60^o = 180^o\\\Rightarrow \angle ADB = 70^o$

Now, $\angle ACB = \angle ADB = 70^{\circ}$

(Angles in the same segment are equal)
$\angle ADB = 70^{\circ} \\ \angle ACB = 70^{\circ}$

Therefore option (C) is correct.