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\(1)\) Prove that in a right triangle, the bisector of right angle halves the angle between the median and the altitude drawn from the same vertex.

\(2)\) Prove that if in a triangle, the ratio of tangents of two angles is equal to the ratio of squares of the sines of these angles, then the triangle is either an isosceles or a right one.

\(3)\) Straight lines \(l_{1},l_{2},l_{3}\) are parallel, \(l_{2}\) lying between \(l_{1}\) and \(l_{3}\) at a distance of \(p\) and \(q\) from them. Find the size of a regular triangle whose vertices lie on the given lines (one on each line).

\(4)\) The diagonal of a rectangle divides its angle in the ratio \(m:n\). Find the ratio of the perimeter of the rectangle to its diagonal.

\(5)\) The acute angle of a parallelogram is equal to \(\alpha\) and the sides are \(a\) and \(b\). Find the tangent of the acute angles formed by the larger diagonal and the sides of the parallelogram.

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## Comments

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TopNewest\[2)~Given:\dfrac{ \tan A}{\tan B}=\dfrac{\sin^2 A}{\sin^2 B}\]

\[Proof:\dfrac{ \frac{\sin A}{\cos A}}{\frac{\sin B}{\cos B}}=\dfrac{\sin^2 A}{\sin^2 B}\]

\[\Rightarrow \sin 2A=\sin 2B\] 2A=2B or 2A=\(\pi\)-2B

Former one gives a isosceles triangle while latter case gives a right angled triangle.

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Here is the answer to the \(4^{th}\) question,

If \(p\) is perimeter and \(d\) is diagonal then-

\(\frac{p}{d}= 2\sqrt{2}\cos\left(\frac{\pi(m-n)}{4(m+n)}\right)\)

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Can you explain your working ?

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Here is the solution-

Let ABCD be a rectangle stated in the question, with \(AB=a\) and \(BC=b\) and \(\angle BAC=\frac{\pi m}{2(m+n)}\) and \(\angle DAC=\frac{\pi n}{2(m+n)}\),

Clearly \(p=2(a+b)\) and \(d=\sqrt{a^2+b^2}\),

\(\frac{p}{d}= \frac{2(a+b)}{\sqrt{a^2+b^2}}\)

Also \(\cos\angle BAC=\frac{a}{\sqrt{a^2+b^2}}\) and \(\cos\angle DAC=\frac{b}{\sqrt{a^2+b^2}}\),

So,

\(\frac{p}{d}= 2(\cos\angle BAC +\cos\angle DAC) \)

I think rest you can do on your own, just place the values of angles and add them by identity

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That was the fastest response I have ever seen on Brilliant! Just wait I am writing my solution.

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Hint for 1): Observe that the median ( with its foot on the hypotenuse ) to the right triangle divides it into \( 2 \) isosceles triangles [ Can you think why ? Does the converse hold true ? ]. Now, apply angle chasing, taking into consideration both the altitude and the angle bisector of right angle.Log in to reply

Hint for 3):Log in to reply