Let m1, m2 be the slopes of two adjacent sides of a square of side a such that a2+11a+3(m12+m22)=220. If one vertex of the square is (10(cosα−sinα),10(sinα+cosα)) , where α∈(0,π2) and the equation of one diagonal is (cosα−sinα)x+(sinα+cosα)y=10, then the value of 72(sin4α+cos4α)+a2−3a+13 is a 3 digit number having digits as

Let A= ( 10 ( cos α − sin α ) , 10 ( sin α + cos α ) ) B D → b e ( cos α − sin α ) x + ( sin α + cos α ) y = 10 ⊥ r d i s tan c e f r o m A t o B D → = a 2 ⇒ a = 10 a 2 + 11 a + 3 ( m 1 2 + m 2 2 ) = 220 ⇒ m 1 2 + m 2 2 = 10 3 ⇒ m 1 = 3 , m 2 = − 1 3 or m 1 = − 3 , m 2 = 1 3 Slope of other diagonal AC = sin α + cos α cos α − sin α A C → i s ( sin α + cos α ) x − ( cos α − sin α ) y = 0 H e n c e sin α + cos α cos α − sin α = tan ( α + π 4 ) ⇒ α = π 6 72 ( sin 4 α + cos 4 α ) + a 2 − 3 a + 13 = 128

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Let $m_{1}, m_{2}$ be the slopes of two adjacent sides of a square of side a such that $a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220$ . If one vertex of the square is $(10 (\cos α - \sin α), 10 (\sin α + \cos α))$ , where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(\cos α - \sin α) x + (\sin α + \cos α) y = 10$ , then the value of $72 (\sin^{4} α + \cos^{4} α) + a^{2} - 3 a + 13$ is a 3 digit number having digits as

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Detailed Solution

Let A= $(10 (\cos α - \sin α), 10 (\sin α + \cos α))$
$\vec{B D} b e (\cos α - \sin α) x + (\sin α + \cos α) y = 10 ⊥^{r} d i s \tan c e f r o m A t o \vec{B D} = \frac{a}{\sqrt{2}} \Rightarrow a = 10 a^{2} + 11 a + 3 (m_{1}^{2} + m_{2}^{2}) = 220 \Rightarrow m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$
$\Rightarrow m_{1} = \sqrt{3,} m_{2} = - \frac{1}{\sqrt{3}}$ or $m_{1} = - \sqrt{3}, m_{2} = \frac{1}{\sqrt{3}}$
Slope of other diagonal AC = $\frac{\sin α + \cos α}{\cos α - \sin α}$
$\vec{A C} i s (\sin α + \cos α) x - (\cos α - \sin α) y = 0 \begin{array}{l} H e n c e \frac{\sin α + \cos α}{\cos α - \sin α} = \tan (α + \frac{π}{4}) \Rightarrow α = \frac{π}{6} \\ 72 (\sin^{4} α + \cos^{4} α) + a^{2} - 3 a + 13 = 128 \end{array}$

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