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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.

Note by Akshat Sharda
10 months, 2 weeks 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. Rishabh Cool · 7 months, 1 week ago

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

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

\(\frac{p}{d}= 2\sqrt{2}\cos\left(\frac{\pi(m-n)}{4(m+n)}\right)\) Akshay Yadav · 7 months, 1 week ago

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@Akshay Yadav Can you explain your working ? Akshat Sharda · 7 months, 1 week ago

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@Akshat Sharda 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 · 7 months, 1 week ago

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@Akshat Sharda That was the fastest response I have ever seen on Brilliant! Just wait I am writing my solution. Akshay Yadav · 7 months, 1 week 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 \( 2 \) 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 · 7 months, 1 week ago

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Hint for 3) :

Karthik Venkata · 7 months ago

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