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Posted by All Subject Nikita Tomar 7 years, 9 months ago
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Posted by Isha Singh 7 years, 9 months ago
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Posted by Isha Singh 7 years, 9 months ago
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Posted by Isha Singh 7 years, 9 months ago
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Tapan Pandey 7 years, 9 months ago
Posted by Prathamesh Katkam 7 years, 9 months ago
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All Subject Nikita Tomar 7 years, 9 months ago
Posted by Tejaswa Jadon 7 years, 9 months ago
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Posted by All Subject Nikita Tomar 7 years, 9 months ago
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All Subject Nikita Tomar 7 years, 9 months ago
Posted by Srushti Bagi 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the distance between pole AB and man be x
{tex}\therefore {/tex} Distance between pole CD and man = 80 - x
In {tex}\triangle {/tex}ABE, {tex}\angle A E B = 30 ^ { \circ }{/tex}
{tex}\tan 30 ^ { \circ } = \frac { h } { x }{/tex}(using Pythagoras theorem)
{tex} \ { \sqrt { 3 } }{/tex} {tex}= \frac { h } { x }{/tex}
{tex}h = \frac { x } { \sqrt { 3 } }{/tex}.........(I)
In {tex}\triangle {/tex}CDE, {tex}\angle C E D = 60 ^ { \circ }{/tex}
{tex}\tan 60 ^ { \circ } = \frac { h } { 80 - x }{/tex}
{tex}\Rightarrow \quad \sqrt { 3 } = \frac { h } { 80 - x }{/tex}
{tex}\Rightarrow \quad h = 80 \sqrt { 3 } - x \sqrt { 3 }{/tex} ....(ii)
Comparing (i) and (ii) we get
{tex}\Rightarrow \quad \frac { x } { \sqrt { 3 } } = 80 \sqrt { 3 } - x \sqrt { 3 }{/tex}
{tex}\Rightarrow \quad x = 80 \times \sqrt { 3 } \times \sqrt { 3 } - x \sqrt { 3 } \times \sqrt { 3 }{/tex}
{tex}\Rightarrow {/tex} 4x = 240
{tex}= \frac { 240 } { 4 } = 60 \mathrm { m }{/tex}
Substituting this value of x in (i)
{tex}h = \frac { 60 } { \sqrt { 3 } } = 20 \sqrt { 3 }{/tex}
Hence, height of the pole = 60m
Distance between pole CD and man = 80 - x
= 80 - 60 = 20m
Posted by All Subject Nikita Tomar 7 years, 9 months ago
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Srushti Bagi 7 years, 9 months ago
Posted by Sujata Malik 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.
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Aakriti Jain 7 years, 9 months ago
0Thank You