Differentiation (Part-38) | S N De

 

$\,1.\,$ In a triangle the sides are in the ratio $\,\sqrt2 :2: (\sqrt3+1).\,\,$ Find the possible values of the angles .

Sol. $\,\frac{a}{\sqrt2}=\frac{b}{2}=\frac{c}{1+\sqrt3}=k(\neq 0),\,\text{say} \\ \therefore  a=\sqrt2k,\,b=2k,\,c=(1+\sqrt3)k.$

So, $\,\cos A \\=\frac{b^2+c^2-a^2}{2bc}\\=\frac{4k^2+(1+\sqrt3)^2k^2-2k^2}{4(1+\sqrt3)k^2}\\=\frac{4+1+3+2\sqrt3-2}{4(1+\sqrt3)}\\=\frac{6+2\sqrt3}{4(1+\sqrt3)}\\=\frac{2\sqrt3(\sqrt3+1)}{4(\sqrt3+1)}\\=\frac{\sqrt3}{2}=\cos 30^{\circ} \\ \Rightarrow A=30^{\circ}$

Again, $\,\cos B\\=\frac{c^2+a^2-b^2}{2ac}\\=\frac{(\sqrt3+1)^2k^2+2k^2-4k^2}{2\sqrt2(\sqrt3+1)k^2}\\=\frac{3+1+2\sqrt3 +2-4}{2\sqrt2(\sqrt3+1)}\\=\frac{2(\sqrt3+1)}{2\sqrt2(\sqrt3+1)}\\=\frac{1}{\sqrt2}=\cos45^{\circ} \\ \Rightarrow B=45^{\circ}.$

Hence, $\,\,C=180^{\circ}-(30^{\circ}+45^{\circ})=105^{\circ}.$

$\,2.\,$ If the angles of a triangle are in the ratio $\,\, 2:3:7\,\,$ and the circumradius is $\,10\,$ cm , find the lengths of its sides.

Sol. Let $\,\,A,B,C\,$ be the angles of the triangle where $\,\,A=2x,\,B=3x,\,C=7x.$

So, $\,\, 2x+3x+7x=180^{\circ} \\ \Rightarrow 12x=180^{\circ}  \\ \Rightarrow x=180^{\circ}/12=15^{\circ}. \\ \therefore A=2\times 15^{\circ}=30^{\circ},\,B=3 \times 15^{\circ}=45^{\circ},\, \\ C=7\times 15^{\circ}=105^{\circ}.$

Now, $\,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \\ \therefore \frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 105^{\circ}}=2\times 10 \\ \Rightarrow \frac{a}{1/2}=\frac{b}{1/\sqrt2}=\frac{c}{\frac{\sqrt3+1}{2\sqrt2}}=20\,[*] \rightarrow(1)$

Note [*] : $\,\sin105^{\circ}\\=\sin(90^{\circ}+15^{\circ})\\=\cos 15^{\circ}\\=\cos(45^{\circ}-30^{\circ})\\=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}\\=(1/\sqrt2)(\sqrt3/2)+(1/\sqrt2)(1/2)\\=\frac{\sqrt3+1}{2\sqrt2}$

Hence, from $\,(1)\,$ we get, $\,a= 10\,\text{cm},\,\,\,b=10\sqrt2\,\,\text{cm}, \\ c=20 \times \frac{\sqrt3+1}{2\sqrt2} \\=5\sqrt2(\sqrt3+1) \\=5(\sqrt6+\sqrt2)\,\,\text{cm.}$

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$\,3.\,$ In $\, \Delta ABC,\,\,\angle B=60^{\circ},\,c=2\sqrt3,\,b=3\sqrt2\,$, then find $\,\angle A.$

Sol. We know, $\,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow \frac{3\sqrt2}{\sin 60^{\circ}}=\frac{2\sqrt3}{\sin C} \\ \Rightarrow \frac{3\sqrt2}{\sqrt3/2}=\frac{2\sqrt3}{\sin C} \\ \Rightarrow \sin C=2\sqrt3 \times \frac{\sqrt3}{2} \times \frac{1}{3\sqrt2}=\frac{1}{\sqrt2} \\ \Rightarrow \sin C=\sin 45^{\circ} \\ \therefore C=45^{\circ} \\ \therefore A=180^{\circ}-(60^{\circ}+45^{\circ})=75^{\circ}.$ 

$\,4.\,$ In $\,\Delta ABC,\,\,\angle A=60^{\circ}\,\,$ and $\,\,b :c=(\sqrt3+1):2.\,\,$ Find the other two angles $\,B\,$ and $\,C.$

Sol.  In $\,\Delta ABC,\,\,\angle A=60^{\circ} \\ \therefore B+C=180^{\circ}-60^{\circ}=120^{\circ}\rightarrow(1)$

Again, $\,\,b:c=(\sqrt3+1):2 \\ \Rightarrow \frac bc=\frac{\sqrt3+1}{2} \\ \Rightarrow \frac{b+c}{b-c}=\frac{\sqrt3+1+2}{\sqrt3+1-2}\,[*] \\ \Rightarrow \frac{2R \sin B+2R\sin C}{2R\sin B-2R\sin C}=\frac{3+\sqrt3}{\sqrt3-1} \\ \left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right] \\ \Rightarrow \frac{\sin B+\sin C}{\sin B-\sin C}=\frac{\sqrt3(\sqrt3+1)}{(\sqrt3-1)} \\ \Rightarrow \frac{2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)}{2\cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)} =\frac{\sqrt3(\sqrt3+1)}{(\sqrt3-1)} \\ \Rightarrow \tan\left(\frac{B+C}{2}\right)\cot\left(\frac{B-C}{2}\right)=\frac{\sqrt3(\sqrt3+1)}{(\sqrt3-1)} \\ \Rightarrow \tan60^{\circ}\cot\left(\frac{B-C}{2}\right)=\frac{\sqrt3(\sqrt3+1)}{(\sqrt3-1)} \\ ~~~~~~~~~~~~~~~~~~~[\because B+C=120^{\circ}] \\ \Rightarrow \sqrt3 \cot\left(\frac{B-C}{2}\right)=\frac{\sqrt3(\sqrt3+1)}{(\sqrt3-1)} \\ \Rightarrow \tan\left(\frac{B-C}{2}\right)=\frac{\sqrt3-1}{\sqrt3+1}=\frac{\tan60^{\circ}-\tan45^{\circ}}{1+\tan60\tan45} \\ \Rightarrow \tan\left(\frac{B-C}{2}\right)=\tan(60^{\circ}-45^{\circ}) \\ \Rightarrow B-C=30^{\circ} \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ we get, $\,\,B=75^{\circ},\,C=45^{\circ}.$

Note[*] : By componendo dividendo formula.

$\,5.\,$ If the two sides and the included angle are respectively $\,a=\sqrt3+1,\,b=2,\,\angle C=60^{\circ},\,\,$ find the other angles and the third side.

Sol. $\,\,\cos C=\frac{b^2+a^2-c^2}{2ab} \\ \Rightarrow \cos 60^{\circ}=\frac{2^2+(\sqrt3+1)^2-c^2}{2(\sqrt3+1)\times 2} \\ \Rightarrow  \frac 12=\frac{4+3+1+2\sqrt3-c^2}{4(\sqrt3+1)} \\ \Rightarrow \frac 12=\frac{8+2\sqrt3-c^2}{4(\sqrt3+1)}\\ \Rightarrow  2(\sqrt3+1)=8+2\sqrt3-c^2 \\ \Rightarrow 2=8-c^2 \\ \Rightarrow  c^2=6 \\ \Rightarrow  c=\sqrt6$

Again, $\,\, \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \therefore  \frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow 2/\sin B=\sqrt6/\sin60^{\circ} \\ \Rightarrow \sin B=2 \times \frac{\sqrt3}{2}\times \frac{1}{\sqrt6} \\ \Rightarrow \sin B=\frac{1}{\sqrt2}=\sin45^{\circ} \\ \therefore B=45^{\circ} \\ \therefore A=180^{\circ}-(60^{\circ}+45^{\circ})=75^{\circ}$

Hence, the other angles are given by $\,\,45^{\circ},\,\,75^{\circ}\,\,$ and the third side is of $\,\,\sqrt6\,\,$ unit.

$\,4.\,$ Find the area of the triangle if $\,\,(i)\,a=13,\,b=14,\,c=15$

Sol. Here, $\,\,s=\frac 12(13+14+15)=\frac{42}{2}=21\,\,\text{unit} \\ \therefore \Delta=\sqrt{s(s-a)(s-b)(s-c)}\,\,\text{sq. unit}\\=\sqrt{21(21-13)(21-14)(21-15)}\,\,\text{sq. unit}\\=\sqrt{21 \times 8\times 7\times 6}\,\,\text{sq. unit}\\=84\,\,\text{sq. unit}$

$\,4.\,$ Find the area of the triangle if $\,\,(ii)\,a:b:c=3:4:5,\,\,s=48\,\,\text{cm}$

Sol. Let $\,\,\,a=3k,\,b=4k,\,c=5k\,\,(k \neq 0) \\ \therefore s=\frac 12(3k+4k+5k)=48 \\ \Rightarrow 6k=48 \\ \Rightarrow k=\frac{48}{6}=8 \\ \therefore a=3k=3\times 8=24\,\text{cm} \\ b=4k=4\times 8=32\,\text{cm} \\ c=5k=5\times 8=40\,\text{cm} \\ \therefore \Delta=\sqrt{s(s-a)(s-b)(s-c)}\,\,\text{sq. unit}\\=\sqrt{48(48-24)(48-32)(48-40)}\,\,\text{sq. unit}\\=\sqrt{48 \times 24\times 16\times 8}\,\,\text{sq. unit}\\=384 \,\,\text{sq. unit}$

$\,4.\,$ Find the area of the triangle if $\,\,(iii)\,a=\left(\frac xy+\frac yz\right),\,b=\left(\frac yz+\frac zx\right),\\ c=\left(\frac zx +\frac xy\right)$

Sol. We have , $\,a=\left(\frac xy+\frac yz\right)\,\,\,\text{unit},\\b=\left(\frac yz+\frac zx\right)\,\,\,\text{unit},\\c=\left(\frac zx +\frac xy\right)\,\,\,\text{unit}. \\ \therefore s=\frac 12(a+b+c)=\left(\frac xy+\frac yz+\frac zx\right)\,\,\,\text{unit} \\ \therefore s-a=\frac zx \,\,\,\text{unit}, \\ s-b=\frac xy \,\,\,\text{unit}, \\ s-c=\frac yz\,\,\,\text{unit}. \\ \therefore \Delta=\sqrt{s(s-a)(s-b)(s-c)}\,\,\,\text{ sq. unit}\\=\sqrt{\left(\frac xy+\frac yz+\frac zx\right) \times \frac zx\times \frac xy\times \frac yz} \,\,\,\text{sq. unit}\\=\sqrt{\frac xy+\frac yz+\frac zx}\,\,\,\text{sq. unit}$