# Questionnaire

Happy Problem solving !!

$1)$ 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.

Note by Akshat Sharda
4 years, 4 months ago

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$2)~Given:\dfrac{ \tan A}{\tan B}=\dfrac{\sin^2 A}{\sin^2 B}$

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

$\Rightarrow \sin 2A=\sin 2B$ 2A=2B or 2A=$\pi$-2B
Former one gives a isosceles triangle while latter case gives a right angled triangle.

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

Here is the answer to the $4^{th}$ question,

If $p$ is perimeter and $d$ is diagonal then-

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

- 4 years, 1 month ago

Can you explain your working ?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Hint for 3) :

- 4 years, 1 month ago