Question

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that \(a^2 + 11a + 3(m_1^2 + m_2^2) = 220\). If one vertex of the square is \((10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))\), where \(\alpha \in \left(0, \frac{\pi}{2}\right)\)and the equation of one diagonal is \((\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10\), then \(72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13\) is equal to :

• A. 119
• B. 128
• C. 145
• D. 155

Answer 1

The Correct Option is B

To solve the problem, we need to determine the expression \(72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13\), given a set of conditions related to a square.

1. From the problem, we have the equation: a^2 + 11a + 3(m_1^2 + m_2^2) = 220.
2. We are given that (\cos \alpha - \sin \alpha) x + (\sin \alpha + \cos \alpha) y = 10 is the equation of the diagonal of the square. The slope of this line is: -\frac{(\cos \alpha - \sin \alpha)}{(\sin \alpha + \cos \alpha)}.
3. The vertex of the square is (10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha)). Simplifying using trigonometric identities:
   • We use identities: \sin^2 \alpha + \cos^2 \alpha = 1, (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha to deduce: (\sin^2 \alpha + \cos^2 \alpha) + 2\sin \alpha \cos \alpha = 1 + 2\sin \alpha \cos \alpha = (\sin \alpha + \cos \alpha)^2.
4. The condition a^2 + 11a + 3(m_1^2 + m_2^2) = 220 means the diagonals of the square intersect at the point given.
5. Using the equation of the diagonal: (\cos \alpha - \sin \alpha) (10(\cos \alpha - \sin \alpha)) + (\sin \alpha + \cos \alpha) (10(\sin \alpha + \cos \alpha)) = 10, solving it yields: 10((\cos \alpha - \sin \alpha)^2 + (\sin \alpha + \cos \alpha)^2) = 10, further simplifies to: 10(1 + 2\sin \alpha \cos \alpha) = 10.
6. From the relation: a^2 + 11a + 220 - a^2 - 11a = 0 \Rightarrow m_1^2 + m_2^2 = \frac{200 - a^2 - 11a}{3}.
7. To find: 72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13, use the identity: (\sin^2 \alpha + \cos^2 \alpha)^2 = \sin^4 \alpha + \cos^4 \alpha + 2\sin^2 \alpha \cos^2 \alpha = 1 to find: 72(\sin^4 \alpha + \cos^4 \alpha) = 18(1 - 2\sin^2 \alpha \cos^2 \alpha) = 72 - 36 \sum.
8. Finally, simplify: 72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13 = 128.

Therefore, the answer is 128.

Answer 2

One vertex of square is
\((10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))\)
and one of the diagonal is
\((\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y = 10\)

So the other diagonal can be obtained as
\((\cos \alpha + \sin \alpha)x - (\cos \alpha - \sin \alpha)y = 0\)

So, point of intersection of diagonal will be
\((5(\cos \alpha - \sin \alpha), 5(\cos \alpha + \sin \alpha))\)

Therefore, the vertex opposite to the given vertex is (0, 0).
So, the diagonal length
\(10\sqrt{2}\)
Side length \((a) = 10\)

It is given that
\(a^2 + 11a + 3(m_1^2 + m_2^2) = 220\)
\(m_1^2 + m_2^2 = 220 - 100 - 110 = 103\)
and \(m_1 \cdot m_2 = -1\)

Slopes of the sides are \(\tan \alpha \quad \text{and} \quad -\cot \alpha\)
\(\tan(2\alpha) = 3 \quad \text{or} \quad \tan(2\alpha) = \frac{1}{3}\)

\(72(\sin 4\alpha + \cos 4\alpha) + a^2 - 3a + 13\)
\(72 \cdot \tan^4 \alpha + \frac{1}{{(1 + \tan^2 \alpha)}^2} + a^2 - 3a + 13 = 128\)
So, the correct option is (B): \(128\)