In right \(\triangle\)BAC,
\(\cot { { 30 }^{ 0 } } =\frac { AC }{ 50 } \)

\(\Rightarrow \quad AC=50\sqrt { 3 } \)
In right \(\triangle\)ACD,
\(\tan { { 60 }^{ 0 } } =\frac { CD }{ 50\sqrt { 3 } } \)
\(\Rightarrow \quad \sqrt { 3 } =\frac { CD }{ 50\sqrt { 3 } } \)
\(\Rightarrow \quad CD=50\sqrt { 3 } \times \sqrt { 3 } =150\quad m\)
\(\Rightarrow\) So the height of hill CD =150 m.

To prove \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) =0 } \) / algebraic sum of deviation from mean is zero
We have,  \(\bar { x } =\frac { 1 }{ n } \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) \)
\(n\bar { x } =\sum _{ i=1 }^{ n }{ { x }_{ i } } \)
Now,  \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\left( { x }_{ 1 }-\bar { x } \right) +\left( { x }_{ 2 }-\bar { x } \right) +.........+\left( { x }_{ n }-\bar { x } \right) \)
\(\Rightarrow \quad \ \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\left( { x }_{ 1 }+{ x }_{ 2 }+.........+{ x }_{ n } \right) -n\bar { x } \)
\(\Rightarrow \quad \ \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\sum _{ i=1 }^{ n }{ { x }_{ i } } -n\bar { x } \)
\(\Rightarrow \quad \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =n\bar { x } -n\bar { x } =0\)
Hence,  \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =0\)

Let 50 % acids in the solution be x and 25 %of other solution be y.
Total volume in the mixture = x + Y
As per question,
x + y = 10                   .....(i)
and
\(\\ \\ \frac { 50 }{ 100 } x+\frac { 25 }{ 100 } y=\frac { 40 }{ 100 } \times 10\)
\(\Rightarrow \quad 2x+y=16\)    .....(ii)
Subtract equation (i) from (ii)
  2x + y = 16
   x + y = 10
\(\underline { -\quad -\quad - } \)
        x = 6
From(i), 6 + y = 10
\(\Rightarrow\)  y = 4
So,  x = 6 and y = 4

Steps of Construction:
1. Draw a line segment AB of 7 cm
2. Taking A and B as centre draw two circle of 3 cm and 2 cm radius

3. Bisect the line AB. Let mid-point of AB is C.
4. Taking C as centre draw a circle of radius AC which will intersect the circle at point P, Q, R and S.
5. Join BP, BQ, AS and AR. These are the required tangents.

\(\triangle\)AOB~\(\triangle\)COD   (AA similarity)
\(\frac { ar(\triangle COD) }{ ar(\triangle AOB) } =\frac { { CD }^{ 2 } }{ AB^{ 2 } } \)
\(=\frac { { CD }^{ 2 } }{ (3CD)^{ 2 } } =\frac { { CD }^{ 2 } }{ 9CD^{ 2 } } =\frac { 1 }{ 9 } \)
ration=1 : 9

Given PQ = 3 cm and PR = 6 cm.
Therefore,  \(\begin{aligned} \frac{\mathrm{PQ}}{\mathrm{PR}} & =\sin \mathrm{R} \\ \end{aligned}\)
or  \(​​​​\begin{aligned} \sin \mathrm{R} & =\frac{3}{6}=\frac{1}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \angle \mathrm{PRQ} & =30^{\circ} \\ \end{aligned}\)
and therefore, \(\begin{aligned} \angle \mathrm{QPR} & =60^{\circ} \end{aligned}\)
You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Here, \(\frac { AB }{ DF } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } ,\frac { BC }{ EF } =\frac { 5 }{ 7.5 } =\frac { 2 }{ 3 } ,\frac { AC }{ DE } =\frac { 3 }{ 4.5 } =\frac { 2 }{ 3 } \)
As, \(\frac { AB }{ DF } =\frac { BC }{ EF } =\frac { AC }{ DE } \)
So, \(\triangle ABC\sim \triangle DFE\)  [by SSS similarity criterion]
Hence, figures (i) and (ii) are similar triangles, but no other pairs of triangles in the given figure are similar.

Here, two dice are thrown, so possible outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total number of outcomes = 36
Let E = Event of getting a multiple of 2 on one die and a multiple of 3 on the other die.
Here, multiples of 2 are 2, 4, and 6, and multiples of 3 are 3 and 6.
So, favourable outcomes for event E are (2, 3), (4, 3), (6, 3), (2, 6), (4, 6), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2) and (6, 4)
Number of favourable outcomes = 11
Required probability = P (E)
= \(\frac { Number \ of \ favourable \ outcomes \ related \ to \ E }{ Total \ number \ of \ outcomes } \)
= \(\frac{11}{36}\)

Given \(\triangle ABC\)in which AB = AC.
To prove BC = 2 CE
Proof: We know that, tangents from an external point to a circle are equal in length.
\(\therefore\)  AD = AF   [A is an external point] ...........(i)
BD = BE   [B is an external point] ...........(ii)
and   CE = CF   [C is an external point] ............(iii)
Now, AB = AC  [given]
\(\Rightarrow\)   AB - AD = AC - AD     [subtracting AD from both sides]
\(\Rightarrow\)   AB - AD = AC - AF    [using Eq. (i)]
\(\Rightarrow\)   BD = CF
\(\Rightarrow\)  BE = CF      [using Eq. (ii)]
\(\Rightarrow\)   BE = CE     [using Eq. (iii)] ........(iv)
Now, BC = CE+BE = CE+CE  [from Eq. (iv)]
 \(\Rightarrow\)   BC = 2CE                                      
Hence proved.

0

(i)Total number of cards in the box
=100+200+50=350
Number of blue cards=50
P(a blue card)\(={50\over 350}={1\over7}\)
(ii)Number of yellow cards=200
Number of non yellow cards =350-200=150
P(not a yellow card)\(={150\over350}={3\over 7}\)
(iii)Number of yellow or blue cards in the box = 200+50=250
Number of neither yellow nor blue cards = 350-250=10
P(neither a yellow nor a blue card)
\(={100\over 350}={2\over 7}\)

\(\left [3\pm \sqrt {19}\over 5\right]\)

23

Given: T BP and TCQ are tangents to the circle whose centre is O. 
Also,  \(\angle \)PBA = 60°
\(\angle \)ACQ = 70°

To determine: \(\angle \)BAC and \(\angle \)BTC
Sol. Join OB and OC
\(\angle \)OBP = 90°
[Tangent makes 90° angle with the radius at the point of contact]
⇒ \(\angle \)OBA + \(\angle \)ABP = 90°
⇒ \(\angle \)1 + 60° = 90° [Given ABP = 60°]
⇒ \(\angle \)1 = 30°       ......(i)
Also  \(\angle \)OCQ = 90°
⇒ OCA + 70° = 90°  ......(i)
⇒ \(\angle \)OCA = 20°   .....(ii)
In OBA ,   OB = OA = radii
⇒ \(\angle \)1 = \(\angle \)4 = \(\angle \)30°     .....(iii)
[ Angles opposite to equal sides of a triangle are equal]
Similarly, In \(\triangle\)OCA
OC = OA
\(\angle \) 5 = 20°​​​​​​​      .....(iv)
From (iii) and (iv)
\(\angle \)BAC = 20°​​​​​​​ + 30°​​​​​​​ = 50°​​​​​​​
⇒  \(\angle \)BOC = 2x50°​​​​​​​ = 100°​​​​​​​
\(\angle \)BOC + BTC = 180°​​​​​​​
100°​​​​​​​ + \(\angle \)BTC = 180°​​​​​​​
⇒ \(\angle \)BTC = 80°​​​​​​​

\(p^{ 2 }+x^{ 2 }+(p^{ 2 }-q^{ 2 })x-q^{ 2 }=0\)
Here a=p²,b=(p²-q²),x= c=q²
D= b²-4ac=(p²-q²)²-4Xp²-4Xp²X(-q²) = (6²+q²)² 
Now  x= \(x=\frac { -b\pm \sqrt { D } }{ 2a } ,\frac { -b-\sqrt { D } }{ 2a } \)
\(\Rightarrow x=\frac { -(p^{ 2 }-q^{ 2 })+\sqrt { (p^{ 2 }+q^{ 2 })^{ 2 } } }{ 2\times p^{ 2 } } ;x=\frac { (-p^{ 2 }-q^{ 2 })-\sqrt { (p^{ 2 }-q^{ 2 })^{ 2 } } }{ 2\times p^{ 2 } } \)
\(\Rightarrow x=\frac { q^{ 2 } }{ p^{ 2 } } ,-1\)