A triangle ABC' is to be constructed such that
\(\frac { CA^{ ' } }{ CA } =\frac { BA^{ ' } }{ BA } =\frac { B^{ ' }C }{ BC } =\frac { 2 }{ 3 } \)
Thi.s means that the ABC 3 triangle ABC is similar to the triangle ABC with scale factor as \(\frac { 2 }{ 3 } \)
Steps of construction: 
Draw a line segment BC = 5 cm
2. With B as centre and radius = AB = 4 cm, draw an arc.
3. With C as centre and radius = AC = 6 cm, draw another arc, meeting the arc drawn in step 2 at the point A.
4. Join AB and AC to obtain  \(\triangle ABC\)
5.. Below BC, make an acute angle \(\angle CBX\)
6.Along BX mark off three points  B₁ ,B_2, , B₃  such that BB₁ = B₁ B₂
7. Join BC
8. From B₂, draw B₂C II B₃C
9. From C, draw CA' II CA, meeting BA at the point A'. Then ABC is the required triangle

Here, \(\angle APB={ 50 }^{ ° }\)
\(\angle PAB=\angle PBA=\frac { { 180 }^{ ° }-{ 50 }^{ ° } }{ 2 } ={ 65 }^{ ° }\)
\(\angle OAB={ 90 }^{ ° }-\angle PAB\)
\(={ 90 }^{ ° }-{ 65 }^{ ° }={ 25 }^{ ° }\)

Putting x = -\(\frac{1}{2}\). in 3x² + 2kx - 3 = 0
\(3\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+2k\left( -\frac { 1 }{ 2 } \right) -3=0\)
⇒ \(\frac{3}{4}\) - k- 3 = 0
⇒ k = \(\frac{3}{4}\)- 3
⇒ k = \(\frac{3-12}{4}\)
⇒ k = \(\frac{-9}{4}\)

\(10\sqrt{3}m\)

86.6 m

line of sight

Yes

Here a= \(\sqrt { 2 } \) , d= \(\sqrt { 8 } -\sqrt { 2 } =\sqrt { 2 } \)
S_n = \(\frac { n }{ 2 } \left[ 2a+(n-1)(d) \right] \)
\(\frac { n }{ 2 } \left[ 2\sqrt { 2 } +(n-1)\sqrt { 2 } \right] \)
\(=\frac { n }{ 2 } \left[ \sqrt { 2 } +\sqrt { 2 } n \right] \)
\(\frac { n(n+1) }{ \sqrt { 2 } } \)

Let kite is at point K 
AK = length of string 
\(\therefore\)  In right \(\Delta\)ABK,  \(\frac { BK }{ AK } =\sin { \theta } \)
\(=\sin { \theta } =\frac { 30 }{ 60 } \Rightarrow \sin { \theta } =\frac { 1 }{ 2 } \Rightarrow \theta ={ 30 }^{ o }\)

OP ⊥ PR 
[ ∵ Tangent and 
radius are ⊥ to each other at the point of contact] 
\(\angle \)OPQ = 90° -50° 
= 40° 
OP = OQ 
[By isosceles triangle's property] 
\(\angle \)OPQ = \(\angle \)OQP = 40°
In \(\triangle\)OPQ, 
⇒ \(\angle \)O + \(\angle \)P + \(\angle \)Q = 180°
⇒ \(\angle \)O + 40° + 40° = 180°
 \(\angle \)O = 180° - 80° = 100°

mx(x-1)+49=0
\(\Rightarrow \) mx²-7mx+49=0
Here a=m,b=-7m,c=49
For equal roots ,D=0
\(\Rightarrow D=b^{ 2 }-4ac\)
\(\Rightarrow 0=(-7m)^{ 2 }-4\times m\times 49\)
\(\Rightarrow 0=49m^{ 2 }-196m\)
\(\Rightarrow 49m^{ 2 }-196m=0\)
\(\Rightarrow 7m(7m-28)=0\)
\(\Rightarrow 7m=0\) or 7m-28 =0
\(\Rightarrow m=0\) or \(m=\frac { 28 }{ 7 } =4\)
but \(m\neq 0\) [ \(\therefore \) In quadratic equation  \(a\neq 0\) ]

Let number of toys be x
\(\therefore \) Cost of one toy = \(\frac { 180 }{ x } \)
Number of toys sold = x-1
Selling price per toy = Rs \(\left( \frac { 180 }{ x } +1 \right) \)
Total SP =CP+profit
ATQ (x-1)  \(\left( \frac { 180 }{ x } +1 \right) \)=180+10
\(\Rightarrow (x-1)\left( \frac { 180 }{ x } +1 \right) =190\)
\(\Rightarrow x^{ 2 }+179x-180=190x\)
\(\Rightarrow x^{ 2 }-11x-180=0\)
\(\Rightarrow (x-20)(x+9)=0\)
\(\Rightarrow x=20,-9\)
\(\Rightarrow x=20\) [Rejected x=-9]

Given: TAS tangent to the circle with centre O at A.
\(\angle \)OBA = 32^o
To find: x
Sol. In \(\triangle\)OAB OA=OB      [RAdii of the same circle]
⇒   \(\angle \)1 = 32^o      [Angle opposite to equal sides of a triangle are equal]

In OAB
\(\angle \)1 + \(\angle \)2 + 32^o = 180^o     [Angle sum property of a triangle]
⇒ \(\angle \)2 + 32^o + 32^o = 180^o ⇒ <2 = 180^o - 64^o = 116^o
\(\angle \)3 = \(1\over2\)\(\angle \)2 = \(1\over2\) x 116^o = 58^o     
TAS is tangent to the circle (Given)
\(\angle \)x = \(\angle \)ABC = 58^o  [Angles in the alternate segments are equal]

Here, we have
a₂ - a₁ = -3.2 - (-1.2) = -3.2 + 1.2 = -2
a₃ - a₂ = -5.2 - (-3.2) = -5.2 + 3.2 = -2
a₄ - a₃ = -7.2 - (-5.2) = -7.2 + 5.2 = -2
and so on.
Since, the difference of any two consecutive terms same. Therefore, the given list of numbers forms a AP and its common difference (d) is - 2.
Now, next three terms of this AP are,
a₅ = a₄ + d = -7.2 + (-2) = -9.2
a₆ = a₅ + d = -9.2 + (-2) = -11.2
and a₇ = a₆ + d = -11.2 + (-2) = -13.2

First term (a) = 3
and common difference (d) = 2nd term - 1st term
= 1-3 = -2

Let altitude of triangle = x. 
∴ hypotenuse of triangle = 2x + 1and base of traiangle = 2x-1
Using Pythagoras theoram,
(2x+1)²=x²+(2x-1)²
\(\Rightarrow 4x^{ 2 }+1+4x=x^{ 2 }+(2x-1)^{ 2 }\)
\(\Rightarrow x^{ 2 }-8x=0\)
\(\Rightarrow x(x-8)=0\)
either x = 0 or x - 8 = 0
Rejecting x = 0,:. x = 8
∴ altitude of triangle = 8 cm
hypotenuse of triangle 2 x 8 + 1 = 17 cm
and base of triangle 2 x 8 - 1 = 15 cm

The sequence is 10, 13, ..... 94.
94 = 10 + (n - 1) 3 \(\Rightarrow\) n = 29
Therefore \(\\ \frac { 29+1 }{ 2 } \) = 15^th term is the middle term
Middle term = 10 + 14 \(\times\) 3 = 52
Sum of first 14 terms = 14 [20 + 13 \(\times\) 3] = 413
Sum of the last 14 terms = \(\frac { 14 }{ 2 } [110+13\times 3]\)
= 1043

Given, OB = 5 cm and radius OA = 3 cm
Use Pythagoras theorem in right \(\Delta\)OAB to find AB
4 cm

Given, TP and TQ are two tangents to a circle and \(\angle\)POQ = 110°. Join PQ.
In \(\triangle OPQ\), OP = OQ  [radii of circle]

\(\Rightarrow\)   \(\angle OPQ=\angle OQP\)    [angles corresponding to equal sides are equal]
Then, \(\angle OPQ=\angle OQP=\frac { { 180 }^{ \circ }-{ 110 }^{ \circ } }{ 2 } =35^{ \circ }\)
Since,  [\(\because\)radius of a circle is perpendicular to the tangent at the point of contact]
\(\therefore \angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\)   \(\angle OPQ=\angle OQP={ 90 }^{ \circ }\)
\(\Rightarrow\)    \(\angle OPQ+\angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\) \({ 35 }^{ \circ }+\angle OPT={ 90 }^{ \circ }\)
\(\Rightarrow\)  \(\angle OPT={ 55 }^{ \circ }\)
\(\Rightarrow\)  \(\angle OQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \(\angle OQP+\angle PQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \({ 35 }^{ \circ }+\angle PQT={ 90 }^{ \circ }\)
\(\Rightarrow\) \(\angle PQT={ 55 }^{ \circ }\)
Now, in \(\triangle PTQ,\quad \angle QPT+\angle PQT+\angle PTQ=180^{ \circ }\)  
[\(\because\)sum of all angles in a triangle is 180°]
\(\Rightarrow\)  \( { 55 }^{ \circ }+{ 55 }^{ \circ }+\angle PTQ={ 180 }^{ \circ }\)
\(\Rightarrow\angle PTQ={ 180 }^{ \circ }-\left( { 55 }^{ \circ }+{ 55 }^{ \circ } \right) ={ 70 }^{ \circ }\quad \)

Here, let us assume that AB and CD be the two towers of heights h m and 30 m respectively.

∠CBD=60^o and ∠ADB=30^o
Consider rt. ∠ed ΔCBD, we obtain
\(\frac { CD }{ BD } ={ tan\quad 60 }^{ o }\)
\(\Rightarrow \quad \frac { 30 }{ BD } =\sqrt { 3 } \Rightarrow BD=\frac { 30 }{ \sqrt { 3 } } \)   ...(i)
Consider rt. ∠ed ΔADB, we obtain
\(\frac { AB }{ BD } ={ tan\quad 30 }^{ o }\)
\(\Rightarrow \quad \frac { h }{ BD } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } h=BD\)   ...(ii)
From (i) and (ii), we have
\(\sqrt { 3 } h=\frac { 30 }{ \sqrt { 3 } } \)
\(\Rightarrow h=\frac { 30 }{ 3 } =10m\)
Also, from (ii), we have BD = \(10\sqrt { 3 } \) m 
Hence, the distance between two towers is \(10\sqrt { 3 } \)  m 
and the height of the other tower is 10 m.

Given: A boy is standing at a distance of 100 m from the bird flying at an elevation of 30^o. A girl is standing on the roof of 20 m high building finds the angle of elevation of the bird to be 45^o Boy and girl are on opposite side of the bird.
To find: Distance between the bird and the girl i.e., BE.
Solution: In ΔACB,
⇒ \(\frac { h }{ 100 } \) = sin 300 ⇒ h = \(\frac { 1 }{ 2 } \) x 100 = 50 m
⇒ BF = h - 20 = (50 - 20) m = 30 m
In ΔBFE,\(\frac { 30 }{ BE } =sin{ 45 }^{ 0 }\Rightarrow \frac { 30 }{ BE } =\frac { 1 }{ \sqrt { 2 } } \Rightarrow 30\sqrt { 2 } \) = BE
BE = 30 x 1.414 = 42.420 = 42.42 m

Let C be the position of the kite and AC be the length of the string which makes an angle of 60° on the ground. The height of the kite from the ground is BC =60 m.
In right angled \(\Delta\)ABC,

\(\begin{aligned} \sin 60^{\circ} & =\frac{P}{H}=\frac{B C}{A C} \\ \end{aligned}\)
\(​​\begin{aligned} \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{60}{A C} \end{aligned}\)     \(\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]\)
\(\begin{aligned} \therefore \quad A C & =\frac{60 \times 2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \end{aligned}\)     [rationalising]
\(\begin{aligned} =\frac{120 \sqrt{3}}{3}=40 \sqrt{3} \mathrm{~m} \end{aligned}\)
Hence, the length of the string is 40\(\sqrt3\)m.

Let the usual speed of plane be x km/h,
\(\frac { 1500 }{ x } -\frac { 1500 }{ x+250 } =\frac { 1 }{ 2 } \)
\(\Rightarrow x^{ 2 }+250x-750000=0\)
\((x+1000)(x-750)=0\)
x = 750
Speed of plane = 750 km/h.
Values depicted here are:
(i) Helping the needy.
(ii) Quick help to the injured.

12 cm

English-12 Sanskrit-16, or English-9 Sanskrit-19

-1275

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

First,draw a circle with centre O and draw a line l. Now, we draw two lines parallel to l, such that one line say m, is a tangent to the cirde and another say n, is a secant to the cirde.

4.3 cm ; 4.3 cm

We know that, in natural numbers, the smallest and the largest number of three-digits which are multiples of 11 are 110 and 900 respectively. Therefore, the sequence of three-digit numbers which are multiples of 11 are 110, 121, 132, ..., 990. Thus, it is an A.P. with first term a = 110, common difference d = 9 and last term (a_n) = l = 990.
Let there are n terms in this sequence.
\(\therefore\)       a_n = 999
\(\Rightarrow\)   a + ( n - 1 )d = 999
\(\Rightarrow\)    110 + ( n - 1)11 = 999
\(\Rightarrow\)       11n = 990 - 99
\(\Rightarrow\)       11n = 891  \(\Rightarrow\) n = 81
Hence, the equired sum = \({81 \over 2}(110+990)\)
\(={81\times1100 \over 2}\)
= 81 x 550 = 44550

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let CD be a tower of height x m and BC is a pole and ㄥBAD=600 and ㄥPCA=450 ㄥPCA=ㄥCAD=45⁰
In right ΔCDA, \(\frac { CD }{ DA } \)=tan 45^o
⇒ \(\frac { x }{ DA } \)=1 ⇒ DA= x m
In right ΔBDA, \(\frac { BD }{ DA } \)=tan 60^o
⇒ \(\frac { 5+x }{ x } =\sqrt { 3 } \)
⇒ 5+x=\(\sqrt { 3 } \)x ⇒ 5=(\(\sqrt { 3 } \)-1)x
⇒ x=\(\frac { 5 }{ \sqrt { 3 } -1 } m=\frac { 5(\sqrt { 3 } +1) }{ 2 } m=\frac { 5(1.73+1) }{ 2 } m=\frac { 13.65 }{ 2 } \)m=6.82 m

Given A\(\Delta ABC\) , in which BC=5cm, \(\angle CAB=120°\) and \(\angle ABC=30°\) .
Then, \(\angle BCA=180°-30°-120°=30°\)