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Sia ? 6 years, 6 months ago
Let speed of the stream be x km/h .
Speed of the boat in still water = 11 km/h (Given)
{tex}\therefore{/tex}Upstream speed of boat =(11 - x)km/h
and, downstream speed be (11+ x) km/h
Distance = 12 km
Also we know that, time taken to cover 'd' km with speed 's' km/h is {tex} =\frac ds{/tex}
Time taken by boat to cover 12 km downstream = {tex}\frac{12}{11 + x}{/tex}hours
Time taken by boat to cover 12 km upstream = {tex}\frac{12}{11 - x}{/tex}hours
ATQ {tex}\frac{12}{11 + x}{/tex} + {tex}\frac{12}{11 - x}{/tex} = {tex}2 \frac { 3 } { 4 }{/tex} {tex}\Rightarrow{/tex} {tex}\frac { 12 ( 11 - x ) + 12 ( 11 + x ) } { ( 11 + x ) ( 11 - x ) }{/tex} = {tex}\frac{11}{4}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { 132 + 132 } { 121 - x ^ { 2 } }{/tex} = {tex}\frac{11}{4}{/tex} {tex}\Rightarrow{/tex} 4 {tex}\times{/tex}264 = 11(121 - x2)
{tex}\Rightarrow{/tex}{tex}\frac{4 \times 264}{11}{/tex} = 121 - x2 {tex}\Rightarrow{/tex} 4 {tex}\times{/tex} 24 = 121 - x2
{tex}\Rightarrow{/tex}x2 = 25 {tex}\Rightarrow{/tex} x = {tex}\pm{/tex}5
Hence,speed of the stream is x = 5 km/h
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Sia ? 6 years, 6 months ago
{tex}x^2 + x - p(p+1) = 0{/tex}
{tex}x^2 + (p+1)x - px - p(p+1) = 0{/tex}
{tex}x(x+ p + 1) -p(x + p + 1) = 0{/tex}
(x + p + 1)(x - p) = 0
x = - p - 1, p
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{tex}L H S = \tan ^ { 2 } A - \tan ^ { 2 } B{/tex}
{tex}= \frac { \sin ^ { 2 } A } { \cos ^ { 2 } A } - \frac { \sin ^ { 2 } B } { \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A \cos ^ { 2 } B - \sin ^ { 2 } B \cos ^ { 2 } A } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A \left( 1 - \sin ^ { 2 } B \right) - \sin ^ { 2 } B \left( 1 - \sin ^ { 2 } A \right) } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A - \sin ^ { 2 } A \sin ^ { 2 } B - \sin ^ { 2 } B + \sin ^ { 2 } B \sin ^ { 2 } A } { \cos ^ { 2 } A \cos ^ { 2 } B }{/tex}
{tex}= \frac { \sin ^ { 2 } A - \sin ^ { 2 } B } { \cos ^ { 2 } A \cos ^ { 2 } B } = R H S{/tex}
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Abc A 7 years, 9 months ago
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