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Sia ? 6 years, 6 months ago
Given A radius OP of a circle C (O, r) and a line APB, perpendicular to OP.
To Prove AB is a tangent to the circle at the point P.
PROOF Take a point Q, different from P, on the line AB.since radius through the point of contact of tangent is perpendicular to it. Therefore,
{tex} O P \perp A B{/tex}.
{tex}{/tex}We know that among all the line segments joining O to a point on AB, OP is the shortest one. Therefore,
{tex}{/tex} OP < <oq>OQ
{tex}{/tex}{tex}\Rightarrow{/tex} OQ > OP</oq>
{tex}\Rightarrow{/tex}Q <oq>lies outside the circle.
Therefore, every point on AB, other than P, lies outside the circle. This implies that AB meets the circle only at the point P.
Hence, AB is a tangent to the circle at P.</oq>
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Sia ? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the speed of the train be x km/hr and that of the car be y km/hr.
Case I Distance covered by train = 250 km.
Distance covered by car = (370 - 250) km = 120 km.
Time taken to cover 250 km by train = {tex}\frac { 250 } { x }{/tex} hours
Time taken to cover 120 km by car = {tex}\frac { 120 } { y }{/tex}hours
Total time taken =4 hours
{tex}\therefore \quad \frac { 250 } { x } + \frac { 120 } { y } = 4 \Rightarrow \frac { 125 } { x } + \frac { 60 } { y } = 2{/tex}.......(i)
Case II Distance covered by train = 130 km.
Distance covered by car = (370 -130) km = 240 km.
Time taken to cover 130 km by train = {tex}\frac { 130 } { x }{/tex} hours
Time taken to cover 240 km by car ={tex} \frac { 240 } { y }{/tex} hours
Total time taken = {tex}4 \frac { 18 } { 60 } \text { hours } = 4 \frac { 3 } { 10 } \text { hours } = \frac { 43 } { 10 }{/tex}hours
{tex}\therefore \quad \frac { 130 } { x } + \frac { 240 } { y } = \frac { 43 } { 10 } \Rightarrow \frac { 1300 } { x } + \frac { 2400 } { y } = 43{/tex}......(ii)
Putting {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex}=v, equations (i) and (ii) become
{tex}125u + 60v = 2{/tex} ... (iii) and {tex}1300u + 2400v = 43{/tex}. ... (iv)
On multiplying (iii) by 40 and subtracting (iv) from the result, we get
5000u - 1300v = 80 - 43 {tex}\Rightarrow{/tex} 3700u = 37
{tex}\Rightarrow u = \frac { 37 } { 3700 } = \frac { 1 } { 100 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 100 } \Rightarrow x = 100{/tex}
Putting u = {tex}\frac{1}{100}{/tex} in (iv), we get
{tex}\left( 1300 \times \frac { 1 } { 100 } \right) + 2400 v = 43 \Rightarrow 2400 v = 43 - 13 = 30{/tex}
{tex}\Rightarrow v = \frac { 30 } { 2400 } = \frac { 1 } { 80 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 80 } \Rightarrow y = 80{/tex}
{tex}\therefore{/tex} x = 100 and y = 80.
Hence, the speed of the train is 100 km/hr and that of the car is 80 km/hr
Posted by Anurag Ojha 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
On dividing n by 3, let q be the quotient and r be the remainder.
Then, {tex}n = 3q + r{/tex}, where {tex}0 \leq r < 3{/tex}
{tex}\Rightarrow\;n = 3q + r{/tex} , where r = 0,1 or 2
{tex}\Rightarrow{/tex} {tex}n = 3q \;or \;n = (3q + 1) \;or\; n = (3q + 2){/tex}.
Case I If n = 3q then n is clearly divisible by 3.
Case II If {tex}\;n = (3q + 1)\; {/tex} then {tex} (n + 2)= (3q + 1 + 2) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.
In this case, {tex}(n + 2){/tex} is divisible by 3.
Case III If n = {tex}(3q + 2){/tex} then {tex}(n + 1) = (3q + 2 + 1) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.
In this case,{tex} (n + 1){/tex} is divisible by 3.
Hence, one and only one out of {tex}n, (n + 1){/tex} and {tex}(n + 2){/tex} is divisible by 3.
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