\displaystyle \textbf{Question 1. }\text{In any }\triangle ABC,\ \sum a^2(\sin B-\sin C)=
\displaystyle \text{(a) }a^2+b^2+c^2\qquad \text{(b) }b^2\qquad \text{(c) }b^2-c^2\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sum a^2(\sin B-\sin C)
\displaystyle =a^2(\sin B-\sin C)+b^2(\sin C-\sin A)+c^2(\sin A-\sin B)
\displaystyle \text{Using sine rule, }a=k\sin A,\ b=k\sin B,\ c=k\sin C
\displaystyle =k^2[\sin^2A(\sin B-\sin C)+\sin^2B(\sin C-\sin A)
\displaystyle +\sin^2C(\sin A-\sin B)]
\displaystyle =(\sin A-\sin B)(\sin B-\sin C)(\sin C-\sin A)
\displaystyle =0
\displaystyle \therefore \text{Correct option is (d).}
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\displaystyle \textbf{Question 2. }\text{In a }\triangle ABC,\text{ if }a=2,\ \angle B=60^\circ\text{ and }\angle C=75^\circ,\text{ then }b=
\displaystyle \text{(a) }\sqrt3\qquad \text{(b) }\sqrt6\qquad \text{(c) }\sqrt9\qquad \text{(d) }1+\sqrt2
\displaystyle \text{Answer:}
\displaystyle A=180^\circ-(60^\circ+75^\circ)=45^\circ
\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}
\displaystyle \therefore b=\frac{a\sin B}{\sin A}
\displaystyle =\frac{2\cdot\frac{\sqrt3}{2}}{\frac{1}{\sqrt2}}=\sqrt6
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 3. }\text{If the sides of a triangle are in the ratio }1:\sqrt3:2,\text{ then the measure of its greatest angle is}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad \text{(b) }\frac{\pi}{3}\qquad \text{(c) }\frac{\pi}{2}\qquad \text{(d) }\frac{2\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Greatest side }=2
\displaystyle \cos A=\frac{1^2+(\sqrt3)^2-2^2}{2\cdot1\cdot\sqrt3}
\displaystyle =\frac{1+3-4}{2\sqrt3}=0
\displaystyle \therefore A=\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 4. }\text{In any }\triangle ABC,\ 2(bc\cos A+ca\cos B+ab\cos C)=
\displaystyle \text{(a) }abc\qquad \text{(b) }a+b+c\qquad \text{(c) }a^2+b^2+c^2\qquad \text{(d) }\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
\displaystyle \text{Answer:}
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}
\displaystyle \therefore 2bc\cos A=b^2+c^2-a^2
\displaystyle \text{Similarly, }2ca\cos B=c^2+a^2-b^2
\displaystyle \text{and }2ab\cos C=a^2+b^2-c^2
\displaystyle \therefore 2(bc\cos A+ca\cos B+ab\cos C)=a^2+b^2+c^2
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 5. }\text{In a triangle }ABC,\ a=4,\ b=3,\ \angle A=60^\circ\text{ then }c\text{ is a root of the equation}
\displaystyle \text{(a) }c^2-3c-7=0\qquad \text{(b) }c^2+3c+7=0
\displaystyle \text{(c) }c^2-3c+7=0\qquad \text{(d) }c^2+3c-7=0
\displaystyle \text{Answer:}
\displaystyle a^2=b^2+c^2-2bc\cos A
\displaystyle 16=9+c^2-2\cdot3\cdot c\cdot\frac{1}{2}
\displaystyle \therefore 16=9+c^2-3c
\displaystyle \therefore c^2-3c-7=0
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 6. }\text{In a }\triangle ABC,\text{ if }(c+a+b)(a+b-c)=ab,\text{ then the measure of angle }C\text{ is}
\displaystyle \text{(a) }\frac{\pi}{3}\qquad \text{(b) }\frac{\pi}{6}\qquad \text{(c) }\frac{2\pi}{3}\qquad \text{(d) }\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle (a+b+c)(a+b-c)=ab
\displaystyle \therefore (a+b)^2-c^2=ab
\displaystyle \therefore a^2+b^2+ab=c^2
\displaystyle \text{Using cosine rule, }c^2=a^2+b^2-2ab\cos C
\displaystyle \therefore a^2+b^2+ab=a^2+b^2-2ab\cos C
\displaystyle \therefore \cos C=-\frac{1}{2}
\displaystyle \therefore C=\frac{2\pi}{3}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 7. }\text{In any }\triangle ABC,\text{ the value of }2ac\sin\left(\frac{A-B+C}{2}\right)\text{ is}
\displaystyle \text{(a) }a^2+b^2-c^2\qquad \text{(b) }c^2+a^2-b^2
\displaystyle \text{(c) }b^2-c^2-a^2\qquad \text{(d) }c^2-a^2-b^2
\displaystyle \text{Answer:}
\displaystyle A+C=\pi-B
\displaystyle \therefore \frac{A-B+C}{2}=\frac{\pi-2B}{2}=\frac{\pi}{2}-B
\displaystyle \therefore \sin\left(\frac{A-B+C}{2}\right)=\cos B
\displaystyle \therefore 2ac\sin\left(\frac{A-B+C}{2}\right)=2ac\cos B
\displaystyle \text{Using cosine rule, }\cos B=\frac{a^2+c^2-b^2}{2ac}
\displaystyle \therefore 2ac\cos B=a^2+c^2-b^2
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 8. }\text{In a }\triangle ABC,\ a(b\cos C-c\cos B)=
\displaystyle \text{(a) }a^2\qquad \text{(b) }b^2-c^2\qquad \text{(c) }0\qquad \text{(d) }b^2+c^2
\displaystyle \text{Answer:}
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}
\displaystyle b\cos C=\frac{a^2+b^2-c^2}{2a}
\displaystyle c\cos B=\frac{a^2+c^2-b^2}{2a}
\displaystyle \therefore a(b\cos C-c\cos B)
\displaystyle =\frac{a^2+b^2-c^2-a^2-c^2+b^2}{2}
\displaystyle =b^2-c^2
\displaystyle \therefore \text{Correct option is (b).}
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