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Naaaaaaaaaaaaaaaaaaa Naaaaaaaaaaaaaaaaaaa 7 years, 9 months ago
Posted by Deepanshu Rawat 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Clearly{tex}\angle O P T{/tex} = 90°
Applying Pythagoras in{tex}\Delta O P T{/tex}, we have
OT2=OP2 +PT2
{tex}\Rightarrow {/tex}132 = 52 + PT2
{tex}\Rightarrow {/tex} PT2 = 169 - 25 = 144
{tex}\Rightarrow {/tex} PT = 12 cm
Since lengths of tangents drawn from a point to a circle are equal. Therefore,
AP = AE = x(say)
{tex}\Rightarrow {/tex} AT = PT - AP = (12 - x)cm
Since AB is the tangent to the circle E. Therefore, {tex}O E \perp A B{/tex}
{tex}\Rightarrow \quad \angle O E A = 90 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \angle A E T = 90 ^ { \circ }{/tex}[Applyng Pythagoras Theorem in {tex}\Delta A E T{/tex}]
{tex}\Rightarrow {/tex} (12 - x)2 = x2 + (13 - 5)2
{tex}\Rightarrow {/tex} 144 - 24x + x2 = x2 + 64
{tex}\Rightarrow {/tex} 24x = 80
{tex}\Rightarrow \quad x = \frac { 10 } { 3 } \mathrm { Cm }{/tex}
Similarly, BE {tex}= \frac { 10 } { 3 } \mathrm { cm }{/tex}
{tex}\therefore{/tex} AB = AE + BE = {tex}\left( \frac { 10 } { 3 } + \frac { 10 } { 3 } \right) \mathrm { cm }{/tex}
{tex}= \frac { 20 } { 3 } \mathrm { cm }{/tex}
Posted by Khursheed Alam 7 years, 9 months ago
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Ayush Agrawal 7 years, 9 months ago
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Ayush Agrawal 7 years, 9 months ago
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Ayush Agrawal 7 years, 9 months ago
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Sia ? 6 years, 5 months ago
Given polynomial is f(x) = x3 - 3x2 + x + 1
Let {tex} \alpha{/tex} = (a - b), {tex} \beta{/tex} = a and {tex} \gamma{/tex} = (a + b)
Now, {tex} \alpha + \beta + \gamma{/tex} = {tex} - \frac { ( - 3 ) } { 1 }{/tex}
⇒ (a - b) + a + ( a + b ) = 3
⇒ a - b + a + a+ b = 3
⇒ a + a + a = 3
⇒ 3a = 3
⇒ a = 3/3
⇒ a = 1
Also, {tex} \alpha \beta + \beta y + \gamma \alpha = \frac { 1 } { 1 }{/tex}
⇒ (a - b)a + a (a + b) + (a + b)(a - b) = 1
⇒ a2 - ab + a2 +ab + a2 - b2 = 1
⇒ 3a2 - b2 = 1 ( ∵ a = 1)
⇒ 3(1)2 - b2 = 1( ∵ a = 1)
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = {tex} \pm \sqrt{2}{/tex}
Hence, a = 1 and b = {tex} \pm \sqrt{2}{/tex}
Posted by Prakash Patel 7 years, 9 months ago
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