\displaystyle \textbf{Question 1. }\text{Find the area of the triangle }\triangle ABC\text{ in which } \\ a=1,\ b=2\text{ and }\angle C=60^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}ab\sin C
\displaystyle =\frac{1}{2}\cdot1\cdot2\cdot\sin60^\circ
\displaystyle =\frac{\sqrt3}{2}
\displaystyle \therefore \text{Area of }\triangle ABC=\frac{\sqrt3}{2}\text{ sq. units.}
\\

\displaystyle \textbf{Question 2. }\text{In a }\triangle ABC,\text{ if }b=\sqrt3,\ c=1\text{ and }\angle A=30^\circ,\text{ find }a.
\displaystyle \text{Answer:}
\displaystyle a^2=b^2+c^2-2bc\cos A
\displaystyle =(\sqrt3)^2+1^2-2\cdot\sqrt3\cdot1\cdot\cos30^\circ
\displaystyle =3+1-2\sqrt3\cdot\frac{\sqrt3}{2}
\displaystyle =4-3=1
\displaystyle \therefore a=1
\\

\displaystyle \textbf{Question 3. }\text{In a }\triangle ABC,\text{ if }\cos A=\frac{\sin B}{2\sin C},\text{ then show that }c=a.
\displaystyle \text{Answer:}
\displaystyle \cos A=\frac{\sin B}{2\sin C}
\displaystyle \therefore 2\sin C\cos A=\sin B
\displaystyle \therefore \sin C\cos A=\frac{\sin B}{2}
\displaystyle \text{Now, }B=\pi-(A+C)
\displaystyle \therefore \sin B=\sin(A+C)
\displaystyle \therefore 2\sin C\cos A=\sin(A+C)
\displaystyle 2\sin C\cos A=\sin A\cos C+\cos A\sin C
\displaystyle \therefore \sin C\cos A=\sin A\cos C
\displaystyle \therefore \tan C=\tan A
\displaystyle \therefore C=A
\displaystyle \therefore c=a
\\

\displaystyle \textbf{Question 4. }\text{In a }\triangle ABC,\text{ if }b=20,\ c=21\text{ and }\sin A=\frac{3}{5},\text{ find }a.
\displaystyle \text{Answer:}
\displaystyle \sin A=\frac{3}{5}
\displaystyle \therefore \cos A=\frac{4}{5}\text{ or }-\frac{4}{5}
\displaystyle a^2=b^2+c^2-2bc\cos A
\displaystyle \text{If }\cos A=\frac{4}{5},\text{ then }a^2=20^2+21^2-2\cdot20\cdot21\cdot\frac{4}{5}
\displaystyle =400+441-672=169
\displaystyle \therefore a=13
\\

\displaystyle \textbf{Question 5. }\text{In a }\triangle ABC,\text{ if }\sin A\text{ and }\sin B\text{ are the roots of the equation } \\ c^2x^2-c(a+b)x+ab=0,\text{ then find }\angle C.
\displaystyle \text{Answer:}
\displaystyle \sin A+\sin B=\frac{a+b}{c}
\displaystyle \text{Using sine rule, }a=k\sin A,\ b=k\sin B,\ c=k\sin C
\displaystyle \therefore \sin A+\sin B=\frac{k\sin A+k\sin B}{k\sin C}
\displaystyle \therefore \sin A+\sin B=\frac{\sin A+\sin B}{\sin C}
\displaystyle \therefore \sin C=1
\displaystyle \therefore C=90^\circ
\\

\displaystyle \textbf{Question 6. }\text{In }\triangle ABC,\text{ if }a=8,\ b=10,\ c=12\text{ and }C=\lambda A,\text{ find the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \frac{\sin C}{\sin A}=\frac{c}{a}=\frac{12}{8}=\frac{3}{2}
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}
\displaystyle =\frac{64+144-100}{2\cdot8\cdot12}=\frac{108}{192}=\frac{9}{16}
\displaystyle \text{Since }B=180^\circ-(A+C),
\displaystyle \sin C=\frac{3}{2}\sin A
\displaystyle \text{This gives }C=2A
\displaystyle \therefore \lambda=2
\\

\displaystyle \textbf{Question 7. }\text{If the sides of a triangle are proportional to }2,\sqrt6\text{ and }\sqrt3-1, \\ \text{ find the measure of its greatest angle.}
\displaystyle \text{Answer:}
\displaystyle \text{Greatest side }=\sqrt6
\displaystyle \text{Let the greatest angle be }A.
\displaystyle \cos A=\frac{2^2+(\sqrt3-1)^2-(\sqrt6)^2}{2\cdot2(\sqrt3-1)}
\displaystyle =\frac{4+(4-2\sqrt3)-6}{4(\sqrt3-1)}
\displaystyle =\frac{2-2\sqrt3}{4(\sqrt3-1)}=-\frac{1}{2}
\displaystyle \therefore A=120^\circ
\displaystyle \therefore \text{Greatest angle }=120^\circ
\\

\displaystyle \textbf{Question 8. }\text{If in a }\triangle ABC,\ \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c},\text{ then find the measures of angles }A,B,C.
\displaystyle \text{Answer:}
\displaystyle \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}
\displaystyle \text{Using sine rule, }a=k\sin A,\ b=k\sin B,\ c=k\sin C
\displaystyle \therefore \frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}=\frac{\cos C}{\sin C}
\displaystyle \therefore \cot A=\cot B=\cot C
\displaystyle \therefore A=B=C
\displaystyle A+B+C=180^\circ
\displaystyle \therefore A=B=C=60^\circ
\\

\displaystyle \textbf{Question 9. }\text{In any triangle }ABC,\text{ find the value of } \\ a\sin(B-C)+b\sin(C-A)+c\sin(A-B).
\displaystyle \text{Answer:}
\displaystyle \text{Using sine rule, }a=k\sin A,\ b=k\sin B,\ c=k\sin C
\displaystyle a\sin(B-C)+b\sin(C-A)+c\sin(A-B)
\displaystyle =k[\sin A\sin(B-C)+\sin B\sin(C-A)+\sin C\sin(A-B)]
\displaystyle =0
\displaystyle \therefore \text{Required value }=0
\\

\displaystyle \textbf{Question 10. }\text{In any }\triangle ABC,\text{ find the value of }\sum a(\sin B-\sin C).
\displaystyle \text{Answer:}
\displaystyle \sum a(\sin B-\sin C)=a(\sin B-\sin C)+b(\sin C-\sin A)+c(\sin A-\sin B)
\displaystyle \text{Using sine rule, }a=k\sin A,\ b=k\sin B,\ c=k\sin C
\displaystyle =k[\sin A(\sin B-\sin C)+\sin B(\sin C-\sin A)+\sin C(\sin A-\sin B)]
\displaystyle =0
\displaystyle \therefore \sum a(\sin B-\sin C)=0
\\

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