Suppose that the digits at units and tens place of the given number be x and y respectively.
Thus, the number is {tex}10y + x.{/tex}
The product of the two digits of the number is 20.
Thus, we have {tex}xy = 20{/tex}
After interchanging the digits, the number becomes {tex}10x + y{/tex}
If 9 is added to the number, the digits interchange their places.
Thus, we have
{tex}(10y + x) + 9 = 10x + y{/tex}
{tex}\Rightarrow{/tex} {tex}10y + x + 9 = 10x+ y{/tex}
{tex}\Rightarrow{/tex} {tex}10x + y - 10y - x = 9{/tex}
{tex}9x - 9y = 9{/tex}
{tex}\Rightarrow{/tex}{tex} 9(x - y) = 9{/tex}
{tex}\Rightarrow x - y = \frac{9}{9}{/tex}
{tex}\Rightarrow{/tex} {tex}x - y = 1{/tex}
So, we have the systems of equations
{tex}xy = 20{/tex} ....(i)
{tex}x - y = 1{/tex} ....(ii)
Here x and y are unknowns.
We have to solve the above systems of equations for x and y.
Substituting {tex}x = 1 + y{/tex} from the second equation to the first equation, we get {tex}(1+ y) y = 20{/tex}
{tex}\Rightarrow{/tex} {tex}y + y^2 = 20{/tex}
{tex}\Rightarrow{/tex} {tex}y^2 + y - 20 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y^2 + 5y - 4y - 20 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y(y + 5) - 4(y + 5) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(y + 5)(y - 4) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}y = -5\ or\ y = 4{/tex}
Substituting the value of y in the second equation, we have

Note that in the first pair of solution the values of x and y are both negative. But the digits of the number can't be negative. So, we must remove this pair.
Hence, the number is 10 {tex}\times{/tex} 4 + 5 = 45

For sin A,
By using identity {tex}cosec ^ { 2 } A - \cot ^ { 2 } A = 1 \Rightarrow \cos e c ^ { 2 } A = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \frac { 1 } { \sin ^ { 2 } A } = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \sin A = \frac { 1 } { \sqrt { 1 + \cot ^ { 2 } A } }{/tex}
For secA, 
By using identity {tex}\sec ^ { 2 } A - \tan ^ { 2 } A = 1 \Rightarrow \sec ^ { 2 } A = 1 + \tan ^ { 2 } A{/tex}
{tex}\Rightarrow \sec ^ { 2 } A = 1 + \frac { 1 } { \cot ^ { 2 } A } = \frac { \cot ^ { 2 } A + 1 } { \cot ^ { 2 } A } \Rightarrow \sec ^ { 2 } A = \frac { 1 + \cot ^ { 2 } A } { \cot ^ { 2 } A }{/tex}
{tex}\Rightarrow \sec A = \frac { \sqrt { 1 + \cot ^ { 2 } A } } { \cot A }{/tex}
For tanA,
{tex}\tan A = \frac { 1 } { \cot A }{/tex}

Given that at the foot of a mountain the elevation of its summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. We have to find the height of the mountain.
Let F be the foot and S be the summit of the mountain FOS. Then {tex}\angle O F S = 45 ^ { \circ }{/tex}and therefore, {tex}\angle O S F = 45 ^ { \circ }.{/tex}Consequently, OF = OS = h km (say). Let FP = 1000 m = 1 km be the slope so that {tex}\angle O F P = 30 ^ { \circ }.{/tex}Draw PM {tex}\perp {/tex}OF. join PS. It is given that {tex}\angle M P S = 60 ^ { \circ }.{/tex}
In {tex}\triangle F P L,{/tex}we have

{tex}\sin 30 ^ { \circ } = \frac { P L } { P F }{/tex}
{tex}\Rightarrow \quad P L = P F \sin 30 ^ { \circ } = \left( 1 + \frac { 1 } { 2 } \right) \mathrm { km } = \frac { 1 } { 2 } \mathrm { km }{/tex}
{tex}\therefore \quad O M = P L = \frac { 1 } { 2 } \mathrm { km }{/tex}
{tex}\Rightarrow \quad M S = O S - O M = \left( h - \frac { 1 } { 2 } \right) \mathrm { km }{/tex} ...(i)
Also, {tex}\cos 30 ^ { \circ } = \frac { F L } { P F }{/tex}
{tex}\Rightarrow \quad F L = P F \cos 30 ^ { \circ } = \left( 1 \times \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km } = \frac { \sqrt { 3 } } { 2 } \mathrm { km }{/tex}
Now, h = OS = OF = OL + LF
{tex}\Rightarrow \quad h = O L + \frac { \sqrt { 3 } } { 2 }{/tex}
{tex}\Rightarrow \quad O L = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }{/tex}
{tex}\Rightarrow \quad P M = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }{/tex} ...(ii)
In {tex}\triangle S P M,{/tex} we have
{tex}\tan 60 ^ { \circ } = \frac { S M } { P M }{/tex}
{tex}\Rightarrow{/tex} SM = PM . tan60 ^° 
{tex}\Rightarrow \quad \left( h - \frac { 1 } { 2 } \right) = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \sqrt { 3 }{/tex}
{tex}\Rightarrow \quad h - \frac { 1 } { 2 } = h \sqrt { 3 } - \frac { 3 } { 2 }{/tex}
{tex}\Rightarrow \quad \sqrt { 3 } h - h = \frac { 3 } { 2 } - \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad h ( \sqrt { 3 } - 1 ) = 1{/tex}
{tex}\Rightarrow \quad h = \frac { 1 } { \sqrt { 3 } - 1 } = \frac { \sqrt { 3 } + 1 } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) } = \frac { \sqrt { 3 } + 1 } { 2 } = \frac { 2.732 } { 2 } = 1.366 \mathrm { km }{/tex}
Hence, the height of the mountain is 1.366 km.

Let a be the first term and d the common difference of the given A.P.
{tex}\therefore S_{p}=\frac{p}{2}{/tex} [2a + (p - 1)d] = q 
{tex}\Rightarrow{/tex} 2a + (p - 1)d {tex}=\frac{2 q}{p}{/tex}  ….(i)
And {tex}S_{q}=\frac{q}{2}{/tex} [2a + (q - 1)d] = p
{tex}\Rightarrow{/tex} 2a + (q - 1)d {tex}=\frac{2 p}{q}{/tex} ….(ii)
Subtracting eq. (ii) from eq. (i) we get
(p - q)d = {tex}\frac{2 q}{p}-\frac{2 p}{q}{/tex} {tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{2\left(q^{2}-p^{2}\right)}{p q}{/tex}
{tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{-2}{p q}{/tex}(p² - q²)
{tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{-2}{p q}{/tex} (p + q)(p - q) {tex}\Rightarrow d=\frac{-2}{p q}{/tex} (p + q)
Substituting the value of d in eq. (i) we get
2a + (p - 1) {tex}\left[\frac{-2(p+q)}{p q}\right]=\frac{2 q}{p}{/tex}
{tex}\Rightarrow 2 a=\frac{2 q}{p}+\frac{2(p-1)(p+q)}{p q}{/tex}
{tex}\Rightarrow a=\frac{q}{p}+\frac{(p-1)(p+q)}{p q}{/tex}
{tex}a=\frac{q^{2}+p^{2}+p q-p-q}{p q}{/tex}
Now S_p+q {tex}=\frac{p+q}{2}{/tex} [2a + (p + q - 1)d
{tex}=\frac{p+q}{2}\left[\frac{2 q^{2}+2 p^{2}+2 p q-2 q-2 q}{p q}+\frac{(p+q-1)[-2(p+q)}{p q}\right]{/tex}
{tex}=\frac{p+q}{2}\left[\frac{2q^{2} + 2p^{2} + 2pq - 2p - 2q -2p^{2} -2 p q+2 p-2 p q-2 q^{2}+2 q}{p q}\right]{/tex}
{tex}=\frac{p+q}{2}\left[\frac{-2 p q}{p q}\right]{/tex} = -(p + q) hence proved.

7 ones + 8 tenth = 7 + 80 = 87
 

a³ + b³ = (a + b) (a²- ab + b²).

The Ancient Greek mathematician Archimedes of Syracuse (287-212 BC) is largely considered to be the first to calculate an accurate estimation of the value of pi.