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PROPERTIES OF TRIANGLES (PART-2)

PROPERTIES OF TRIANGLES (PART-2)

 

$\,\,7.\,$ In $\,\,\Delta ABC,\,a=3,\,b=5,\,c=7,\,\,$ show that , the triangle is obtuse angled.

Sol. $\,\, \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{9+25-49}{2\times 3\times 5} \\ \Rightarrow \cos C=\frac{-15}{30}=-\frac 12 \\ \Rightarrow \cos C=\cos 120^{\circ} \\ \Rightarrow C=120^{\circ}$

Hence, the triangle is obtuse angled.

$\,8.\,$ If $\,\,\frac{\sin A}{3}=\frac{\sin B}{3}=\frac{\sin C}{4},\,\,$ then prove that, $\,\,\cos C=\frac 19.$

Sol. $\,\,\frac{\sin A}{3}=\frac{\sin B}{3}=\frac{\sin C}{4},\,\,\text{(Given)} \\ \therefore \frac a3=\frac b3=\frac c4=k(\neq 0) \\ \left[\because  \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\right] \\ \therefore a=3k,\,b=3k,\,c=4k. \\ \therefore \cos C=\frac{(3k)^2+(3k)^2-(4k)^2}{2.3k.3k} \\ \Rightarrow \cos C=\frac{18k^2-16k^2}{18k^2}=\frac{2k^2}{18k^2}=\frac 19.$

$\,9.\,$ If in a triangle ABC, $\,\,\frac{a-b+c}{a}=\frac{b}{b+c-a}\,\,$ then prove that $\,\,\angle C=60^{\circ}.$ 

Sol. We have, $\,\,\frac{a-b+c}{a}=\frac{b}{b+c-a} \\ \Rightarrow (a-b+c)(b+c-a)=ab \\ \Rightarrow ab+ac-a^2-b^2-bc+ab+bc \\ +c^2-ac=ab \\ \Rightarrow ab=a^2+b^2-c^2 \rightarrow(1)$

Now, $\,\,\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{ab}{2ab}\,\,[\text{By (1)}] \\ \Rightarrow \cos C=\frac 12=\cos 60^{\circ} \\ \Rightarrow C=60^{\circ}\,\,\text{(proved)}$

To download full PDF  S N De Math solution of PROPERTIES OF TRIANGLES, click here .

$\,10.\,\,$ If $\,\,a=3,\,b=4,\,c=5,\,$ find the value of $\,\,\tan(B/2).$

Sol. Here, $\,s=\frac 12(a+b+c)=\frac{3+4+5}{2}=6 \\ \therefore \tan \frac B2=\frac{(s-c)(s-a)}{\Delta} \\ \Rightarrow \tan(B/2)\\=\frac{(6-5)(6-3)}{\sqrt{6(6-3)(6-4)(6-5)}}\\=\frac{3}{\sqrt{6\times 3\times 2\times 1}}\\=\frac{3}{6} =\frac 12 \,\,\text{(ans.)}$

$\,11.\,\,$ If $\,a=2b,\,\,A=3B,\,\,$ find the angles of the $\,\,\triangle ABC.$

Sol. We know, $\,\, \frac{a}{\sin A}=\frac{b}{\sin B} \\ \Rightarrow \frac{2b}{\sin 3B}=\frac{b}{\sin B} \\ \Rightarrow \frac{\sin3B}{\sin B}=2 \\ \Rightarrow \frac{3\sin B-4\sin^3B}{\sin B}=2 \\ \Rightarrow 3-4\sin^2B=2 \\ \Rightarrow \sin^2B=\frac 14 \\ \Rightarrow \sin B=\frac 12 \,\,[\because 0^{\circ}<B<180^{\circ}] \\ \therefore B=30^{\circ} \\ \therefore A=3\times 30^{\circ}=90^{\circ}, \\ C=180^{\circ}-(30^{\circ}+90^{\circ})=60^{\circ}.$

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