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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 the 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: $\angle Q P R=60^{\circ}$
Let $\angle Q O R=x$
As we know the angle between the tangent and radius of a circle is 90

$
\angle P Q O=\angle P R O=90^{\circ}
$
We know that $\angle P Q O+\angle P R O+\angle Q P R+\angle Q O R=360^{\circ}$
[ $\because$. the sum of interior angles of a quadrilateral is $360^{\circ}$ ]

$
\begin{aligned}
& 90^{\circ}+90^{\circ}+x+60^{\circ}=360^{\circ} \\
& 240+x=360^{\circ} \\
& x=120^{\circ}
\end{aligned}
$
 

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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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3. To construct a triangle similar to a given $\triangle A B C$ with its sides of 3/7 the corresponding sides $\triangle A B C$, first, draw a ray BX such that $C B X$ is an acute angle and $X$ lies on the opposite side of $A$ with respect to $B C$. Then locate points $B_1, B_2, B_3, \ldots$ on $B X$ at equal distances and the 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.

In order to construct a triangle similar to a given $\triangle A B C$ with its sides $3 / 7$ we have to divide $B C$ in the ratio $3: 7$
$B X$ should have 7 equidistant points on it as 7 is a greater number
Now we have to join $B_7$ to $C$.

Therefore, the next step is to join B7 to C.

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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}\cdotsand B_{1},B_{2},B_{3}\cdotsare 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 a distance equal to the distance between points on the AX line.
5. Join A_{5}B_{6}.
Let it intersect AB at a point C in the 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}             
\thereforeby 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

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 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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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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In Fig. 10.9, \angleAOB = 90º and \angleABC = 30º, then \angleCAO is equal to:

(A) 30º

(B) 45º

(C) 90º

(D) 60º

In AOB,

\angleOAB +\angleABO + \angleBOA = 180°      … (i)                           (angle sum property of Triangle)

OA = OB = radius

Angles opposite to equal sides are equal

\angleOAB = \angleABO

Equation (i) becomes

\angleOAB + \angleOAB + 90° = 180°

2\angleOAB = 180° – 90°

\angleOAB  = 45° …(ii)

In \triangleACB,

\angleACB + \angleCBA + \angleCAB = 180°   (angle sum property of Triangle)

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

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

\therefore  45° + 30° + \angleCAB = 180°

\angleCAB = 180° – 75° = 105°

\angleCAO + \angleOAB = 105°

\angleCAO + 45° = 105°

\angleCAO = 105° – 45° = 60°

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In Fig. 10.8, BC is a diameter of the circle and \angleBAO = 60º. Then \angleADC is equal to:

Fig. 10.8

(A) 30º

(B) 45º

(C) 60º

(D) 120º

In \triangleAOB, AO = OB    (Radius)

\angleABO = \angleBAO         [Angle opposite to equal sides are equal]

\angleABO = 60° [\therefore \angleBAO = 60° Given]

In \triangleAOB,

\angleABO + \angleOAB =\angleAOC (exterior angle is equal to the sum of interior opposite angles)

 60° + 60° = 120° =\angleAOC

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

(The 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 )

\angleADC = 60°

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ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and \angleADC = 140º, then \angleBAC is equal to:

(A) 80º

(B) 50º

(C) 40º

(D) 30º

Given, ABCD is cyclic Quadrilateral and \angleADC = 140°

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

\angleADC + \angleABC = 180°

140° + \angleABC = 180°

\angleABC = 180° – 140°

\angleABC = 40°

Since \angleACB is an angle in a circle

\angleACB = 90°

In \triangleABC

\angleBAC + \angleACB + \angleABC = 180° (angle sum property of a triangle)

\angleBAC + 90° + 40° = 180°

\angleBAC = 180° – 130° = 50°

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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

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}

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