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# 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
1 year, 2 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. · 11 months ago

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

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

$$\frac{p}{d}= 2\sqrt{2}\cos\left(\frac{\pi(m-n)}{4(m+n)}\right)$$ · 11 months ago

Can you explain your working ? · 11 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 · 11 months ago

That was the fastest response I have ever seen on Brilliant! Just wait I am writing my solution. · 11 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. · 11 months, 1 week ago

Hint for 3) :

· 11 months ago