Differentiation (Part-38) | S N De

 

$\,9.\,$ In any $\,\,\triangle ABC,\,\,$ if $\,\,\sin A : \sin B: \sin C=4: 5: 6,\,\,$ then prove that, $\,\cos A : \cos B : \cos C=12 : 9: 2.$

Sol. We know, for any  $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC$

Let $\,\, \sin A=4x,\,\sin B=5x,\,\sin C=6x \\ \therefore \text{by (1) we get, } \,\,a=2Rx \times 4\\~~~~~b=2Rx \times 5 \\ ~~~~~c=2Rx \times 6 \\ \therefore \cos A\\=\frac{b^2+c^2-a^2}{2bc}\\=\frac{(2Rx)^2(5^2+6^2-4^2)}{2(2Rx)^2(5 \times 6)}\\= \frac{25+36-16}{ 2\times 30}\\=\frac 34$

Similarly, $\,\,\cos B\\=\frac{c^2+a^2-b^2}{2ca}\\=\frac{(2Rx)^2(6^2+4^2-5^2)}{2(2Rx)^2 (4 \times 6)}\\=\frac{36+16-25}{2 \times 24}\\= \frac{9}{16}$

Finally, in a similar way we can get, $\,\,\cos C= \frac 18$

Hence, $\,\,\cos A : \cos B: \cos C\\=\frac 34 : \frac{9}{16} : \frac 18\\=12 :9 : 2$

$\,10.\,$ In any $\,\triangle ABC,\,$ if $\,\,\tan(A/2),\,\tan(B/2),\,\tan(C/2)\,\,$ are in A.P. , prove that , $\,\,\cos A,\,\cos B,\,\cos C$ are also in A.P.

Sol. By question, $\,\,\tan(B/2)-\tan(A/2)=\tan(C/2)\\ -\tan(B/2) \\ \Rightarrow \frac{\sin(B/2)}{\cos(B/2)}-\frac{\sin(A/2)}{\cos(A/2)}=\frac{\sin(C/2)}{\cos(C/2)}-\frac{\sin(B/2)}{\cos(B/2)}\\ \Rightarrow \frac{\sin\left(\frac{B-A}{2}\right)}{\cos(B/2)\cos(A/2)} =\frac{\sin\left(\frac{C-B}{2}\right)}{\cos(C/2)\cos(B/2)} \\ \Rightarrow \sin\left(\frac{B-A}{2}\right)\cos(C/2)\\ =\sin\left(\frac{C-B}{2}\right)\cos(A/2)\\ \Rightarrow 2\sin\left(\frac{B-C}{2}\right)\cos(A/2)\\ =2\sin\left(\frac{A-B}{2}\right)\cos(C/2)\\ \Rightarrow \sin\left(\frac{B-C+A}{2}\right)+\sin\left(\frac{B-C-A}{2}\right)\\=\sin\left(\frac{A-B+C}{2}\right)+\sin\left(\frac{A-B-C}{2}\right)\\ \Rightarrow \sin \left(\frac{B-C+A}{2}\right)-\sin\left(\frac{A-B+C}{2}\right)\\=\sin\left(\frac{A-B+C}{2}\right)-\sin\left(\frac{B+C-A}{2}\right)\\ \Rightarrow \sin\left(\frac{\pi-2C}{2}\right)-\sin\left(\frac{\pi-2B}{2}\right)\\=\sin\left(\frac{\pi-2B}{2}\right)-\sin\left(\frac{\pi-2A}{2}\right) \\ \Rightarrow \sin\left(\pi/2-C\right)-\sin\left(\pi/2-B\right)\\=\sin\left(\pi/2-B\right)-\sin\left(\pi/2-A\right)\,\,\,[*] \\ \Rightarrow \cos C-\cos B=\cos B-\cos A \rightarrow(1)$

Hence, from $\,(1)\,$ we can conclude that $\,\,\cos A,\,\cos B,\,\cos C$ are also in A.P.

Note[*] : $\, A+B+C=\pi$

$\,11.\,$ If $\,h\,$ be the length of the perpendicular drawn from $\,A\,$ on $\,BC\,$ in $\,\,\triangle ABC\,$, show that $\,\,h=\frac{a \sin B\sin C}{\sin(B+C)}$

Sol. We know, for any  $\,\triangle ABC,\,\, \frac{a}{\sin A}=\frac{c}{\sin C}\rightarrow(1)$,where $\,R\,$ being the circumradius $\,\,\triangle ABC$

By question, $\,\,h=c \sin B \rightarrow(2)\\ \therefore \frac{a\sin B\sin C}{\sin(B+C)}\\=\frac{(a \sin C)\sin B}{\sin(\pi-A)} \\=\frac{c \sin A \sin B}{\sin A}\,\,[\text{By (1)}]\\=c\sin B\rightarrow(3)$

Hence, from $\,(2),\,(3)\,\,$ the result follows.

$\,12.\,$ If $\,D\,$ is the mid-point of the side $\,BC\,$ of $\,\,\triangle ABC\,\,$ and $\,\angle BAD=\theta,\,\,\angle CAD=\psi,\,\,$ then show that, $\,\,\cot\theta-\cot\psi=\cot B-\cot C$

Sol. $\,\,\cot B-\cot C\\=\frac{c^2+a^2-b^2}{4\Delta}-\frac{a^2+b^2-c^2}{4\Delta}\,[*]\\=\frac{2(c^2-b^2)}{4\Delta}\\=\frac{c^2-b^2}{2\Delta}$

Note [*] : We know, $\,\,\tan B=\frac{abc}{R}\frac{1}{c^2+a^2-b^2}=\frac{4\Delta}{c^2+a^2-b^2} \\ \Rightarrow \cot B=\frac{c^2+a^2-b^2}{4\Delta}$ 

Now, for $\,\,\triangle ABC,\,\, AB=c, \,BC=a,\,CA=b,\\ AD=h\,\text{(say)}$

So, $\, \triangle ABD=\frac 12.BD.h \\ \Rightarrow \frac 12\Delta ABC=\frac 12.(a/2).h \\ \Rightarrow \Delta ABC=\frac 12.ah \\ \Rightarrow \Delta=(c\sin\theta)h \,\,\,[\because \frac a2=c\sin\theta] \\ \Rightarrow \sin\theta=\frac{\Delta}{ch}$

Now, $\,\,\cot\theta\\=\frac{\cos\theta}{\sin\theta}\\=\frac{AB^2+AD^2-BD^2}{2AB.AD} \times \frac{1}{\Delta/ch}\\=\frac{c^2+h^2-a^2/4}{2ch} \times \frac{ch}{\Delta}\\=\frac{c^2+h^2-a^2/4}{2\Delta}$

Again, $\,\,\triangle ACD=\frac 12.(a/2)h \\ \Rightarrow \frac 12 \triangle ABC=\frac 12(b \sin\psi)h\,\,[\because \frac a2=b\sin\psi] \\ \Rightarrow \triangle ABC=\sin\psi.bh \\ \therefore \sin\psi=\frac{\Delta}{bh}$

Similarly, we can deduce $\,\cot\psi=\frac{b^2+h^2-a^2/4}{2\Delta}$

Hence, $\,\,\cot \theta-\cot\psi\\=\frac{c^2+h^2-a^2/4}{2\Delta}-\frac{b^2+h^2-a^2/4}{2\Delta}\\=\frac{c^2-b^2}{2\Delta}\\=\cot B-\cot C\,\,\text{(showed)}$

$\,13.\,$ The sides of a triangle are $\,4,\,5,\,6\,$ cm . Show that its smallest angle is half of its greatest angle.

Sol. Let $\,ABC\,\,$ be a triangle with sides $\,a=4\,\text{cm},\,b=5\,\,\text{cm},\,c=6\,\,\text{cm}$

So,$\,\,\cos A\\=\frac{b^2+c^2-a^2}{2bc}\\=\frac{5^2+6^2-4^2}{2.5.6}\\=\frac 34\rightarrow(1)\\ \cos C\\=\frac{a^2+b^2-c^2}{2ab}\\=\frac{4^2+5^2-6^2}{2.4.5}\\=\frac 18\rightarrow(2) $

Now, $\,\,\cos 2A\\=2\cos^2A-1\\=2\times \left(\frac 34\right)^2-1\\=\frac 18 \rightarrow(3)\\ \therefore \cos 2A=\cos C\,\,[\text{By (2),(3)}] \\ \Rightarrow 2A=C$

Hence, the smallest angle of the given triangle is half of its greatest angle.

$\,\,14.\,\,ABC\,$ is a triangle and $\,D\,$ is the middle point of $\,BC.\,$ If $\,AD \perp AC,\,$ prove that $\,\, \cos A\cos C=\frac{2(c^2-a^2)}{3ca}$

Sol. In $\,\,\triangle  ADC,\,\angle DAC=90^{\circ}. \\ \therefore \cos C=\frac{AC}{DC}=\frac{b}{a/2}=\frac{2b}{a} \rightarrow(1) \\ \therefore \cos A \cos C\\=\frac{b^2+c^2-a^2}{2bc}.\frac{2b}{a} \\=\frac{b^2+c^2-a^2}{ac}\rightarrow(2)$ 

Again, from $\,(1),\,$ we get 

$\,\,\frac{a^2+b^2-c^2}{2ab}=\frac{2b}{a} \\ \Rightarrow \frac{a^2+b^2-c^2}{2b}=2b \\ \Rightarrow a^2+b^2-c^2=4b^2 \\ \Rightarrow a^2-c^2=3b^2 \\ \Rightarrow b^2=\frac{a^2-c^2}{3}\rightarrow(3)$ 

From $\,(2),\,(3)\,\,$ we get, 

$\,\,\cos A\cos C \\=\frac{1}{ac}\left[\frac{a^2-c^2}{3}+c^2-a^2\right]\\=\frac{1}{ac}\left[\frac 23(c^2-a^2)\right]\\=\frac{2(c^2-a^2)}{3ac}\,\,\text{(proved)}$

$\,15.\,\,$ In any $\,\,\triangle ABC,\,$ if $\,\,\frac{\cos A+2\cos C}{\cos A+2\cos B}=\frac{\sin B}{\sin C}\,\,$ then prove that, the triangle is either isosceles or right angled.

Sol. $\,\,\frac{\cos A+2\cos C}{\cos A+2\cos B}=\frac{\sin B}{\sin C}\\ \Rightarrow \cos A \sin C+2\sin C\cos C\\=\cos A\sin B+2\sin B\cos B \\ \Rightarrow \cos A(\sin C-\sin B)+\sin2C\\-\sin 2B=0 \\ \Rightarrow \cos A(\sin B-\sin C)\\+\sin 2B-\sin2C=0 \\ \Rightarrow 2\cos A \cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)\\+2\cos(B+C)\sin(B-C)=0 \\ \Rightarrow 2\cos A\cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)\\+2\cos(\pi-A)\times 2\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{B-C}{2}\right)=0 \\ \Rightarrow 2\cos A\cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)\\-2\cos A\times 2\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{B-C}{2}\right)=0 \\ \Rightarrow 2\cos A\sin\left(\frac{B-C}{2}\right)\left[\cos\left(\frac{B+C}{2}\right)\\-2\cos\left(\frac{B-C}{2}\right)\right]=0 \\ \Rightarrow 2\cos A \sin\left(\frac{B-C}{2}\right) =0\\~~ [\because  \frac{B-C}{2}< 90^{\circ},\,\cos\left(\frac{B+C}{2}\right)\\-2\cos\left(\frac{B-C}{2}\right)\neq 0] \\ \Rightarrow A=90^{\circ}\\ \text{or,}\,\,B=C \rightarrow(1)$

Hence, from $\,(1),\,$ we can conclude that the triangle is either isosceles or right angled.