Questionnaire

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1)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)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)3) Straight lines l1,l2,l3l_{1},l_{2},l_{3} are parallel, l2l_{2} lying between l1l_{1} and l3l_{3} at a distance of pp and qq from them. Find the size of a regular triangle whose vertices lie on the given lines (one on each line).

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

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

Note by Akshat Sharda
3 years, 11 months ago

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2) Given:tanAtanB=sin2Asin2B2)~Given:\dfrac{ \tan A}{\tan B}=\dfrac{\sin^2 A}{\sin^2 B}

Proof:sinAcosAsinBcosB=sin2Asin2BProof:\dfrac{ \frac{\sin A}{\cos A}}{\frac{\sin B}{\cos B}}=\dfrac{\sin^2 A}{\sin^2 B}

sin2A=sin2B\Rightarrow \sin 2A=\sin 2B 2A=2B or 2A=π\pi-2B
Former one gives a isosceles triangle while latter case gives a right angled triangle.

Rishabh Jain - 3 years, 8 months ago

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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 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.

Karthik Venkata - 3 years, 8 months ago

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Here is the answer to the 4th4^{th} question,

If pp is perimeter and dd is diagonal then-

pd=22cos(π(mn)4(m+n))\frac{p}{d}= 2\sqrt{2}\cos\left(\frac{\pi(m-n)}{4(m+n)}\right)

Akshay Yadav - 3 years, 8 months ago

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Can you explain your working ?

Akshat Sharda - 3 years, 8 months ago

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Here is the solution-

Let ABCD be a rectangle stated in the question, with AB=aAB=a and BC=bBC=b and BAC=πm2(m+n)\angle BAC=\frac{\pi m}{2(m+n)} and DAC=πn2(m+n)\angle DAC=\frac{\pi n}{2(m+n)},

Clearly p=2(a+b)p=2(a+b) and d=a2+b2d=\sqrt{a^2+b^2},

pd=2(a+b)a2+b2\frac{p}{d}= \frac{2(a+b)}{\sqrt{a^2+b^2}}

Also cosBAC=aa2+b2\cos\angle BAC=\frac{a}{\sqrt{a^2+b^2}} and cosDAC=ba2+b2\cos\angle DAC=\frac{b}{\sqrt{a^2+b^2}},

So,

pd=2(cosBAC+cosDAC)\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 - 3 years, 8 months ago

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

Akshay Yadav - 3 years, 8 months ago

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Hint for 3) :

Karthik Venkata - 3 years, 7 months ago

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