Questionnaire

Please provide solutions if possible.

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.

#Geometry #Questionnaire

Easy Math Editor

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Comments

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

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

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

Rishabh Jain - 5 years, 3 months 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.

Venkata Karthik Bandaru - 5 years, 3 months ago

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

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

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

Akshay Yadav - 5 years, 3 months ago

Can you explain your working ?

Akshat Sharda - 5 years, 3 months 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

Akshay Yadav - 5 years, 3 months ago

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

Akshay Yadav - 5 years, 3 months ago

Hint for 3) :

Venkata Karthik Bandaru - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$