In {tex}\triangle A B C,{/tex} {tex}\tan 60 ^ { \circ } = \frac { A B } { B C }{/tex} 
{tex}\Rightarrow \quad A B = \sqrt { 3 } B C{/tex}  .....(i)
In {tex}\triangle A B D,{/tex}  
{tex}\tan 30 ^ { \circ } = \frac { A B } { B C + 40 }{/tex} 
{tex}\frac{ 1}{√3}=\frac{AB}{BC+40}=\frac{√3BC}{BC+40}{/tex}
3BC = BC + 40
BC = 20, Hence from (i) we get
AB = 20√3 = 20 {tex}\times{/tex} 1.73 = 34.6 meter

A graph of p(x) cut y-axis
∴ At y-axis x = 0
∴ p(0) = a(0)² + b(0) + c

= 0 + 0 + c
= c

{tex}\Rightarrow\ p(0)=c{/tex}
Now

If p(x) cut y-axis in positive direction
∴ p(0) > 0  so that c>0.
Therefore, c is positive

Suppose x and y be two cars starting from points A and B respectively.
Let the speed of the car X be x km/hr and that of the car Y be y km/hr.
Case I: When two cars move in the same directions:
Suppose two cars meet at point Q, then,
Distance travelled by car X = {tex}AQ{/tex}
Distance travelled by car Y = {tex}BQ{/tex}
Distance travelled by car X in 8 hours = 8x km.
{tex}AQ = 8x{/tex}
Distance travelled by car Yin 8 hours = 8y km.
{tex}BQ = 8y{/tex}
Clearly {tex}AQ - BQ = AB{/tex}
{tex}8x - 8y = 80{/tex}
{tex}\Rightarrow {/tex} {tex}x - y = 10{/tex} ...(i)
Case II: When two cars move in opposite direction
Suppose two cars meet at point P, then,
Distance travelled by X car X = AP
Distance travelled by Y car Y = BP
we can write it as 1 hour = {tex}\frac{{20}}{{60}}{/tex}hours.
Therefore, Distance travelled by a car Y in {tex}\frac{4}{3}{/tex}hrs = {tex}\frac{4}{3}{/tex}x km.
AP + BP = AB
{tex}\frac{4}{3}x{/tex} km + {tex}\frac{4}{3} y{/tex} km = 80
{tex}\frac{4}{3}{/tex}{tex}(x + y) = 80{/tex}
{tex}( x + y ) = 80{/tex} × {tex}\frac{3}{4}{/tex}
{tex}x + y = 60{/tex} ...(ii)

By solving equations (i) and (ii), we get, {tex}x = 35.{/tex}
By substituting x = 35 in equation (ii), we get
{tex}x + y = 60{/tex}
{tex}35 + y = 60{/tex}
{tex}y = 60 - 35{/tex}
{tex}y = 25.{/tex}
Hence, speed of car X {tex}= 35\ km/hr.{/tex}
Speed of car Y ={tex} 25\ km/hr.{/tex}

Perimeter means distance around a figure or curve. We can only measure perimeter of a closed figure/2 dimensional shape or curve as movement around a closed figure or curve is possible.

Perimeter of a square = sum of all sides = 4 x side

Perimeter of Rectangle = L + L + B + B

Smallest angle = 120degree

Common difference = 5

A P is 120, 125, 130,……..

The sum of interior angles of a polygon =  (n - 2)180

Hence Sum of n terms of an A P = (n - 2)180

n/2 {2.120+(n-1)5} = 180(n-2)

5n ²  - 125n + 720 = 0

n² -25n + 144 = 0

n = 9 or 16

Hence number of sides can be 9 or 16.

Let son age =x

So father age=x^{​​​​​​2}

Before one year

x^{​​​​​​2}​​​​​-1=8*(x-1)

x2-8x+7=0

(x-7)(x-1)=0 , X=7 or 1

son 7 yr father 49 yr

{tex}\;\frac\theta{360\;}\times{/tex}2 πr = 44
Putting r = 42cm, we get= 60°
Now, Area of minor segmen t= Area of minor sector-Area of ∆
Since = 60°, so the triangle formed will be an equilateral ∆
Area of minor segment = Area of minor sector-Area of equilateral ∆
i.e. Area of minor segment = {tex}\;\frac\theta{360\;}\times\mathrm{πr}^2\;-\;\frac{\sqrt3}4\mathrm a^{{}^2}{/tex}

(-4) + (-1) + 2 + 5 + ---- + x = 437.
Now,
-1 - (-4) = -1 + 4 = 3
2 - (-1) = 2 + 1 = 3
5 - 2 = 3
Thus, this forms an A.P. with a = -4, d = 3,l = x
Let their be n terms in this A.P.
Then,
S_n = {tex}\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] {/tex}
{tex}\Rightarrow 437 = \frac { n } { 2 } [ 2 \times ( - 4 ) + ( n - 1 ) \times 3 ]{/tex}
{tex}\Rightarrow{/tex} 874 = n[-8 + 3n - 3]
{tex}\Rightarrow{/tex}874 = n[3n - 11]
{tex}\Rightarrow{/tex}874 = 3n² - 11n
{tex}\Rightarrow{/tex}3n² - 11n - 874 = 0
{tex}\Rightarrow{/tex}3n² - 57n + 46n - 874 = 0
{tex}\Rightarrow{/tex}3n(n - 19) + 46(n - 19) = 0
{tex}\Rightarrow{/tex}3n + 46 = 0 or n = 19
{tex}\Rightarrow n = - \frac { 46 } { 3 }{/tex} or n = 19
Numbers of terms cannot be negative or fraction.
{tex}\Rightarrow{/tex} n = 19
Now, S_n = {tex}\frac { n } { 2 } [ a + l ]{/tex}
{tex}\Rightarrow 437 = \frac { 19 } { 2 } [ - 4 + x ]{/tex}
{tex}\Rightarrow - 4 + x = \frac { 437 \times 2 } { 19 }{/tex}
{tex}\Rightarrow - 4 + x = 46{/tex}
{tex}\Rightarrow x = 50{/tex}

Secants And Tangents

A secant is a line that intersects the circle in two different points and a tangent is a line that intersects the circle in exactly one point, called the point of tangency. Secant and tangent theorems can be used to find congruency, similarity, and special length relationships between the two. One important theorem about secants and tangents states that the measure of an angle formed by two secants, a secant and a tangent, or two tangents intersecting in the interior of a circle is equal to one-half the difference of the measures of the intercepted arcs; that is, .

Let ABC be the right triangle right angled at A
whose sides AB and AC measure 3 cm
and 4 cm respectively,
{tex}\therefore \text { hypotenuse } \mathrm { BC } = \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } = 5 \mathrm { cm }{/tex}

Let AO (A'O) is the radius of the common base of the double cone formed by revolving the triangle around BC.
Now, the area of {tex}\triangle A B C = \frac { 1 } { 2 } \times 4 \times 3 = \frac { 1 } { 2 } \times B C \times A O{/tex}
{tex}\Rightarrow 6 = A 0 \times \frac { 5 } { 2 } \Rightarrow A O = \frac { 12 } { 5 } = 2.4 \mathrm { cm }{/tex}
Now, the volume of the double cone
{tex}= \frac { 1 } { 3 } \times \pi \times \mathrm { AO } ^ { 2 } \times \mathrm { BO } + \frac { 1 } { 3 } \times \pi \times \mathrm { A } \mathrm { O } ^ { 2 } \times \mathrm { CO }{/tex}
{tex}= \frac { 1 } { 3 } \pi \times A O ^ { 2 } \times ( B O + C O ){/tex}
{tex}= \frac { 1 } { 3 } \times \pi \times \mathrm { AO } ^ { 2 } \times \mathrm { BC } [ \because \mathrm { BO } + \mathrm { CO } = \mathrm { BC } ]{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 2.4 \times 2.4 \times 5 = 30.17 \mathrm { cm } ^ { 3 }{/tex}
The surface area of the double cone = {tex}\pi \times A O \times A B + \pi \times A O \times A C{/tex}
{tex}= \pi \times A O [ A B + A C ]{/tex}
{tex}= \frac { 22 } { 7 } \times 2.4 \times ( 3 + 4 ) = 22 \times 2.4 = 52.8 \mathrm { cm } ^ { 2 }{/tex}