Differentiation (Part-38) | S N De

 

$\,1.\,$ The sides of a $\,\triangle ABC\,\,$ are $\,(x^2+x+1),\,(2x+1),\,(x^2-1).\,$ Find the greatest angle of the triangle. 

Sol. $\,\,a=x^2+x+1, \\ b=2x+1,\\ c=x^2-1. \\ \cos A=\frac{b^2+c^2-a^2}{2bc} \\ \therefore \cos A=\frac{(2x+1)^2+(x^2-1)^2-(x^2+x+1)^2}{2(2x+1)(x^2-1)}\,[**] \\ \Rightarrow \cos A=-\frac 12=\cos 120^{\circ} \\ \Rightarrow A=120^{\circ} \\ \therefore B+C=180^{\circ}-120^{\circ}=60^{\circ}$

So, clearly the greatest angle of the triangle is $\,120^{\circ}.$

Note[**] : $\,(2x+1)^2+(x^2-1)^2\\-(x^2+x+1)^2 \\=(2x+1)^2+(x^2-1+x^2\\+x+1)(x^2-1-x^2-x-1)\\=(2x+1)^2+(2x^2+x)(-x-2)\\=(2x+1)^2-x(2x+1)(x+2)\\=(2x+1)[2x+1-x(x+2)]\\=(2x+1)[2x+1-x^2-2x]\\=-(2x+1)(x^2-1)$

$\,2.\,$ In a $\,\triangle ABC,\,$ if $\,(a^2+b^2)\sin(A-B)=(a^2-b^2)\sin(A+B)\,\,(\text{where}\,\,a \neq b),\,$ prove that the triangle is right angled.

Sol.  $\,(a^2+b^2)\sin(A-B)=(a^2-b^2)\sin(A+B)\\ \Rightarrow \frac{a^2+b^2}{a^2-b^2}=\frac{\sin(A+B)}{\sin(A-B)} \\ \Rightarrow\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}=\frac{\sin(A+B)+\sin(A-B)}{\sin(A-B)-\sin(A-B)} \\ \Rightarrow \frac{2a^2}{2b^2}=\frac{2\sin A\cos B}{2\cos A \sin B}\\ \Rightarrow\frac{a^2}{b^2}=\frac{\sin A\cos B}{\sin B\cos A}\\ \Rightarrow\frac{a^2}{b^2}=\frac{\frac{a}{2R}.\frac{a^2+c^2-b^2}{2ac}}{\frac{b}{2R}.\frac{b^2+c^2-a^2}{2bc}}\\ \Rightarrow\frac{a^2}{b^2}=\frac{a^2+c^2-b^2}{b^2+c^2-a^2}\\ \Rightarrow a^2(b^2+c^2-a^2)=b^2(a^2+c^2-b^2) \\ \Rightarrow a^2b^2+a^2c^2-a^4=a^2b^2+b^2c^2-b^4\\ \Rightarrow c^2(a^2-b^2)=a^4-b^4 \\ \Rightarrow c^2=\frac{(a^2+b^2)(a^2-b^2)}{a^2-b^2} \\ \Rightarrow c^2=a^2+b^2 \rightarrow(1)$

Hence, by $\,(1),\,$ we can conclude that the triangle is right angled.

$\,3.\,$ If $\,\,a^4+b^4+c^4+a^2b^2=2c^2(a^2+b^2)\,\,$ then show that $\,\,C=60^{\circ}\,\,\text{or}\,\, 120^{\circ}.$

Sol. $\,\,a^4+b^4+c^4+a^2b^2=2c^2(a^2+b^2)\,\,\text{(given)} \\ \Rightarrow a^4+b^4+c^4+2a^2b^2-2c^2a^2 \\ -2c^2b^2=a^2b^2 \\ \Rightarrow (a^2+b^2-c^2)^2=a^2b^2 \\ \Rightarrow (a^2+b^2-c^2)=\pm ab \\ \Rightarrow \frac{a^2+b^2-c^2}{2ab}=\pm\frac 12 \\ \Rightarrow \cos C=\pm\frac 12 \\ \therefore C= 60^{\circ}\,\,\text{or}\,\,\, 120^{\circ}$

$\,4.\,$ A wire of length $\,2\,$ metre is cut into three pieces so as to form the sides of a triangle. If the two angles of the triangle be $\,35^{\circ}\,$ and $\,85^{\circ}\,$, find in metre the lengths of the sides. Given $\sin 35^{\circ} = 0.5736,\, \sin 60^{\circ} = 0.8660\,\,$ and $\,\sin 85^{\circ} = 0.9962$.

Sol. Let $\,A=35^{\circ},\,B=85^{\circ} \\ \therefore C=180^{\circ}-35^{\circ}-85^{\circ}=60^{\circ}.$

Now, $\,\,a=2R \sin A=2R \sin 35^{\circ}=2R \times 0.5736 \\ b=2R\sin B=2R \sin 85^{\circ}=2R\times 0.9962 \\ c=2R \sin C=2R \sin 60^{\circ}=2R \times 0.8660$

By question, $\,\,a+b+c=2 \\ \Rightarrow 2R(0.5736+0.9962+0.8660)=2 \\ \Rightarrow R=\frac{2}{2.4358} \\ \therefore a= \frac{2}{2.4358} \times 0.5736=0.47 \\ b=\frac{2}{2.4358} \times 0.9962=0.71 \\ c=\frac{2}{2.4358} \times 0.8660=0.82$

$\,5.\,\,$ The angles $\,A,B,C\,$ of $\,\triangle ABC\,\,$ are in A.P. and $\,b :c= \sqrt3 : \sqrt2.\,\,$ Find $\, A.$

Sol. We have,  $\,b :c= \sqrt3 : \sqrt2.$

Since the angles $\,A,B,C\,$ of $\,\triangle ABC\,\,$ are in A.P.,

 $\, 2B=A+C \\ \therefore \sin(2B)=\sin(A+C)\\ \Rightarrow 2\sin B\cos B=\sin A\cos C+\cos A\sin C \\ \Rightarrow 2. \frac{b}{2R}.\frac{a^2+c^2-b^2}{2ac}=\frac{a}{2R}.\frac{a^2+b^2-c^2}{2ab}\\ +\frac{b^2+c^2-a^2}{2bc}.\frac{c}{2R} \\ \Rightarrow b. \frac{a^2+c^2-b^2}{ac}=\frac{a^2+b^2-c^2}{2b}  +\frac{b^2+c^2-a^2}{2b} \\ \Rightarrow b. \frac{a^2+c^2-b^2}{ac}=\frac{a^2+b^2-c^2+b^2+c^2-a^2}{2b} \\ \Rightarrow b. \frac{a^2+c^2-b^2}{ac}=\frac{2b^2}{2b} \\ \Rightarrow b. \frac{a^2+c^2-b^2}{ac}=b \\ \Rightarrow  \frac{a^2+c^2-b^2}{2ac}=\frac 12 \\ \Rightarrow \cos B=\frac 12 \\ \Rightarrow B=60^{\circ}$

Again, $\,\frac{b}{c}=\frac{\sqrt3}{\sqrt2}\,\,\text{(given)} \\ \Rightarrow \frac{\sin B}{\sin C}=\frac{\sqrt3}{\sqrt2} \\ \Rightarrow \sin C=\frac{\sqrt2}{\sqrt3}\times \sin 60^{\circ}=\frac{\sqrt2}{\sqrt3} . \frac{\sqrt3}{2} \\ \Rightarrow \sin C=\frac{1}{\sqrt2}=\sin 45^{\circ} \\ \Rightarrow C=45^{\circ}$

So, $\,\,A=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}.$

$\,6.\,$ If in $\,\,\triangle ABC ,\,\,\,\tan^{-1} \left(\frac{a}{b+c}\right)+\tan^{-1}\left(\frac{c}{a+b}\right)=\pi/4,\,\,$ then prove that , the triangle is right angled.

 Sol. $\,\,\tan^{-1} \left(\frac{a}{b+c}\right)+\tan^{-1}\left(\frac{c}{a+b}\right)=\pi/4\\ \Rightarrow \tan^{-1}\left[\frac{\frac{a}{b+c}+\frac{c}{a+b}}{1-\frac{a}{b+c}.\frac{c}{a+b}}\right]=\pi/4 \\ \Rightarrow \tan^{-1} \left[\frac{a(a+b)+c(b+c)}{(b+c)(a+b)-ac}\right]=\pi/4 \\ \Rightarrow \frac{a^2+ab+bc+c^2}{ab+b^2+ca+bc-ac}=\tan (\pi/4) \\ \Rightarrow \frac{a^2+ab+bc+c^2}{ab+b^2+bc}=1 \\ \Rightarrow a^2+ab+bc+c^2=ab+b^2+bc \\ \Rightarrow a^2+c^2=b^2 \rightarrow(1)$

From $\,\,(1)\,\,$ we can conclude that the triangle is right angled.

$\,7.\,\,$ If $\, \frac{\sin A}{\sin C}=\frac{\sin(A-B)}{\sin(B-C)},\,\,$ then show that , $\,a^2,\,b^2,\,c^2\,$ are in A.P.

Sol. $\,\,\frac{\sin A}{\sin C}=\frac{\sin(A-B)}{\sin(B-C)} \\ \Rightarrow \frac{\sin B}{\sin C}=\frac{\sin A\cos B-\cos A\sin B}{\sin B\cos C-\cos B\sin C} \\ \Rightarrow  \frac ac=\frac{a.\frac{a^2+c^2-b^2}{2ac}-b.\frac{b^2+c^2-a^2}{2bc}}{b.\frac{a^2+b^2-c^2}{2ab}-\frac{a^2+c^2-b^2}{2ac}.c}\,\,[*] \\ \Rightarrow \frac ac=\frac{\frac{1}{2c}(a^2+c^2-b^2-b^2-c^2+a^2)}{\frac{1}{2a}.(a^2+b^2-c^2-a^2-c^2+b^2)}\\ \Rightarrow  \frac ac=\frac ac. \frac{2(a^2-b^2)}{2(b^2-c^2)}\\ \Rightarrow  1=\frac{a^2-b^2}{b^2-c^2}\\ \Rightarrow  b^2-c^2=a^2-b^2\rightarrow(1)$

Hence, from $\,(1)\,\,$ we can conclude that $\,a^2,\,b^2,\,c^2\,$ are in A.P.

Note[*]: We know, for any  $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\,\,$,where $\,R\,$ being the circumradius $\,\,\triangle ABC$

$\,8.\,$ If $\,a,\,b,c\,$ are in A.P. prove that 

$\,(i)\,\,\cot(A/2),\,\cot(B/2),\,\cot(C/2)\,\,$ are in A.P. 

$\,(ii)\,\,3\tan(A/2)\tan(C/2)=1$

Sol. $\,(i)\,\,\cot(A/2)+\cot(C/2)\\=\frac{s(s-a)}{\Delta}+\frac{s(s-c)}{\Delta}\\=\frac{2s^2-s(a+c)}{\Delta}\\=\frac{2s^2-2bs}{\Delta}\,\,[\because 2b=a+c]\\=2.\frac{s(s-b)}{\Delta}\\=2\cot(B/2)$

Sol. $\,(ii)\,\, 3\tan(A/2)\tan(C/2)\\=3.\frac{(s-b)(s-c)(s-a)(s-b)}{\Delta^2}\\=3.\frac{(s-a)(s-c)(s-b)^2}{s(s-a)(s-b)(s-c)}\\=3.\frac{s-b}{s}\\=3.\frac{\frac{a+b+c}{2}-b}{\frac{a+b+c}{2}}\\=3.\frac{a+c-b}{a+c+b}\\=3.\frac{2b-b}{2b+b}\,\,[\because a+c=2b]\\=3.\frac{b}{3b}\\=1\,\,\,\text{(proved)}$