Question

If the sum of length of the hypotenuse and a side of a right angled triangle is given, then show that if the area of triangle is maximum, then the angle between them is $\frac{\pi }{3}$.

Answer 1

Let ABC be a triangle with AC = h AB = x and BC = y.
Also, $\angle CAB=\theta$
Let h +x = k

$\therefore cos\theta =\frac{x}{h}\Rightarrow x=hcos\theta \Rightarrow h+hcos\theta =k\Rightarrow h(1+cos\theta )=k\Rightarrow h=\frac{k}{(1+cos\theta )}Also,Areaof△ABC=\frac{1}{2}(AB.BC)A=\frac{1}{2}.x.y=\frac{1}{2}hcos\theta hsin\theta =\frac{1}{2}{h}^{2}sin\theta cos.cos\theta =\frac{2h^2}{4}sin\theta .cos\theta =\frac{1}{4}{h}^{2}sin2\theta Sinceh=\frac{k}{1+cos\theta }\therefore A=\frac{1}{4}\left(\frac{k}{1+cos\theta }\right).sin2\theta \Rightarrow A=\frac{k^2}{4}.\frac{sin2\theta }{(1+cos\theta {)}^2}\therefore \frac{dA}{d\theta }=\frac{k^2}{4}\left[\frac{(1+cos\theta {)}^2.cos2\theta 2-isn2\theta 2(1+cos\theta ).(0-sin\theta )}{(1+cos\theta {)}^4}\right]=\frac{k^2}{4}\left\{\frac{2(1+cos\theta )\left[\right(1+cos\theta ).cos2\theta +sin2\theta (sin\theta \left)\right]}{(1+coscos\theta {)}^4}\right\}=\frac{k^2}{4}.\frac{2}{(1+cos\theta {)}^3}\left[\right(1+cos\theta ).cos2\theta +2si{n}^2\theta .cos\theta ]=\frac{k^2}{2(1+cos\theta {)}^2}\left[\right(1+cos\theta \left)\right(1-2si{n}^2\theta )+si{n}^2\theta .cos\theta ]=\frac{k^2}{2(1+cos\theta {)}^3}[1+cso\theta -2si{n}^2\theta -2si{n}^2\theta ]=\frac{k^2}{2(1+cos\theta {)}^3}\left[\right(1+cos\theta )-2si{n}^2\theta ]=\frac{k^2}{2(1+cos\theta {)}^3}[1+cos\theta -2+2co{s}^2\theta ]=\frac{k^2}{2(1+cos\theta {)}^3}(co{s}^2\theta +cos\theta -1)for\frac{dA}{d\theta }=0,\frac{k^2}{2(1+cos\theta {)}^2}(2co{s}^2\theta +cos\theta -1)=0\Rightarrow 2co{s}^2\theta +cos\theta -1=0\Rightarrow 2co{s}^2\theta +2cos\theta -cos\theta -1=0\Rightarrow 2cos\theta (cos\theta +1)-1(cos\theta +1)=1\Rightarrow (2cos\theta -1)(cos\theta +1)=0\Rightarrow cos\theta =\frac{1}{2}orcos\theta =-1\Rightarrow \theta =\frac{\pi }{3}or\theta =2n\pi \pm \pi \therefore \theta =\frac{\pi }{3}$
Again, differentiating w.r.t. $\theta$ in eq. (v) . we get
$\frac{d}{d\theta }\left(\frac{dA}{d\theta }\right)=\frac{d}{d\theta }\left[\frac{k^2}{2(1+cos\theta )}\right(2co{s}^2\theta +cos\theta -1\left)\right]\therefore \frac{d^{2}A}{d{\theta }^2}=\frac{d}{d\theta }\left[\frac{k^2(cos\theta -1)(1+cos\theta )}{2(1+cos\theta {)}^2}\right]=\frac{d}{d\theta }[\frac{k^2}{2}.\frac{(2cos\theta -1)}{(1+cos\theta {)}^2}]=\frac{k^2}{2}\left[\frac{(1+cos\theta {)}^2.(-2sin\theta )-2(1+cos\theta ).(-sin\theta \left)\right(2cos\theta -1)}{(1+cos\theta {)}^4}\right]=\frac{k^2}{2}\left[\frac{(1+cos\theta ).[1+cos\theta ](-2sin\theta )+2sin\theta (2cos\theta -1)}{(1+cos\theta {)}^4}\right]\frac{k^2}{2}\left[\frac{-2sin\theta -2sin\theta .cos\theta +4sin\theta .cos\theta -2sin\theta }{(1+cos\theta {)}^3}\right]\frac{k^2}{2}\left[\frac{-4sin\theta -sin2\theta +2sin2\theta }{(1+cos\theta {)}^3}\right]=\frac{k^2}{2}\left[\frac{sin2\theta -4sin\theta }{(1+cos\theta {)}^3}\right]{\left(\frac{d^{2}A}{d{\theta }^2}\right)}_{at\theta =\frac{\pi }{3}}=\frac{k^2}{2}\left[\frac{sin\frac{2\pi }{3}-4sin\frac{\pi }{3}}{(1+cos\frac{\pi }{3})}\right]=\frac{k^2}{2}\left[\frac{\frac{\sqrt{3}}{2}-\frac{4\sqrt{3}}{2}}{(1+\frac{1}{2})}\right]=\frac{k^2}{2}\left[\frac{-3\sqrt{3}.8}{2.27}\right]=-k^2\left(\frac{2\sqrt{3}}{9}\right)$
Which is less than zero.
Hence, area of the right angled triangle is maximum when the angle between them is $\frac{\pi }{3}$