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#10.1

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) $120^{\circ}$

Answer(D) 120°
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 $360^{\circ}$ ]

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

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#10.1

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

Answer (B) 8
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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#10.1

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

Answer(C) B₇ 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 B₇to C

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

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#10.1

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

Answer(A) A₅ and B₆
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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#10.1

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

Answer(B) A₁₁
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.

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#10.1

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

Answer(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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#10.1

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)

OA = OB = radius

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.

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#10.1

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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#10.1

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.

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#10.1

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

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.

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All Questions

To draw a pair of tangents to a circle which are inclined to each other at an Angle of , it is required to draw tangents at end points of those two radii of The circle, the angle between them should be (A) (B) (C) (D)

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To construct a triangle similar to a given with its sides of the corresponding sides of draw a ray BX such that 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

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To divide a line segment AB in the ratio , draw a ray AX such that is an acute angle, then draw a ray BY parallel to AX and the points and 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

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JEE Main high-scoring chapters and topics

To divide a line segment AB in the ratio , a ray AX is drawn first such that is an acute angle and then points are located at equal distances on the ray AX and the point B is joined to: (A) A12 (B) A11 (C) A10 (D) A9

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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) $120^{\circ}$

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

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

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

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

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

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