In any $\triangle ABC$, prove the following:

Question 1: In $\triangle ABC$, if $a = 5, b = 6$ and $C = 60^o$, show that the area is $\frac{15\sqrt{3}}{2}$  sq. units.

The area of a triangle is given by  $\triangle =$ $\frac{1}{2}$ $ab \sin C$

Therefore Area $=$ $\frac{1}{2}$ $\times 5 \times 6 \times \sin 60^o =$ $\frac{15\sqrt{3}}{2}$ sq units.

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Question 2: In $\triangle ABC$, if $a = \sqrt{2}, b = \sqrt{3}$ and $C = \sqrt{5}$, show that the area is $\frac{\sqrt{6}}{2}$  sq. units.

The area of a triangle is given by  $\triangle =$ $\frac{1}{2}$ $ab \sin C$

$\cos C =$ $\frac{a^2 + b^2 - c^2}{2 ab}$ $=$ $\frac{2+3-5}{2\sqrt{6}}$ $= 0$

$\sin C = \sqrt{1 - \cos^2 C} = 1$

$\therefore Area =$ $\frac{1}{2}$ $ab \sin C =$ $\frac{1}{2}$ $\times \sqrt{2} \times \sqrt{3} \times 1 =$ $\sqrt{ \frac{3}{2} }$

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Question 3: The sides of a triangle are $a = 4, b = 6$ and $c = 8$, show that: $8 \cos A + 16 \cos B + 4 \cos C = 17$.

Given $a = 4, b = 6$ and $c = 8$

$\cos A =$ $\frac{b^2 + c^2 - a^2}{2 bc}$ $=$ $\frac{36+64-16}{2 \times 6\times 8}$ $=$ $\frac{7}{8}$

$\cos B =$ $\frac{a^2 + c^2 - b^2}{2 ac}$ $=$ $\frac{16+64-36}{2 \times 4\times 8}$ $=$ $\frac{11}{16}$

$\cos C =$ $\frac{a^2 + b^2 - c^2}{2 ab}$ $=$ $\frac{16+36-64}{2 \times 4\times 6}$ $= -$ $\frac{1}{4}$

LHS $= 8 \cos A + 16 \cos B + 4 \cos C$

$= 8 \Big($ $\frac{7}{8}$ $\Big) + 16 \Big($ $\frac{11}{16}$ $\Big) + 4 \Big( -$ $\frac{1}{4}$ $\Big)$ $= 7 + 11 - 1 = 17 =$ RHS. Hence proved.

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Question 4: In $\triangle ABC$, if $a = 18, b = 24, c = 30$, find $\cos A, \cos B$ and $\cos C$.

Given $a = 18, b = 24$ and $c = 30$

$\cos A =$ $\frac{b^2 + c^2 - a^2}{2 bc}$ $=$ $\frac{576+900-324}{2 \times 24\times 30}$ $=$ $\frac{1152}{1440}$ $=$ $\frac{4}{5}$

$\cos B =$ $\frac{a^2 + c^2 - b^2}{2 ac}$ $=$ $\frac{324+900-576}{2 \times 18\times 30}$ $=$ $\frac{648}{1080}$ $=$ $\frac{3}{5}$

$\cos C =$ $\frac{a^2 + b^2 - c^2}{2 ab}$ $=$ $\frac{324+576-900}{2 \times 18\times 24}$ $=$ $\frac{0}{864}$ $= 0$

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Question 5: $b ( c \cos A - a \cos C) = c^2 - a^2$

RHS $= c^2 - a^2$

$= k^2 \sin^2 C - k^2 \sin^2 A$

$= k^2 (\sin^2 C - \sin^2 A)$

$= k^2 \sin ( C+A) \sin(C-A)$

$= k^2 \sin ( \pi - B) \sin(C-A)$

$= bk \sin ( C-A)$

$= bk ( \sin C \cos A - \cos C \sin A)$

$= b ( k\sin C \cos A - \cos C k\sin A)$

$= b ( c \cos A - a \cos C) =$ LHS. Hence proved.

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Question 6: $c ( a \cos B - b \cos A) = a^2 - b^2$

LHS $= c ( a \cos B - b \cos A)$

$= ac \cos B - bc \cos A$

$= ac \Big($ $\frac{a^2 + c^2 - b^2}{2 ac}$ $\Big) - bc \Big($ $\frac{b^2 + c^2 - a^2}{2 bc}$ $\Big)$

$=$ $\frac{1}{2}$ $\Big( a^2 + c^2 - b^2 - b^2 - c^2 + a^2 \Big)$

$=$ $\frac{1}{2}$ $\Big( 2a^2 - 2b^2 \Big)$

$= a^2 - b^2 =$ RHS. Hence proved.

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Question 7: $2( bc \cos A + ca \cos B + ab \cos C) = a^2 + b^2 + c^2$

LHS $2( bc \cos A + ca \cos B + ab \cos C)$

$= 2bc \Big($ $\frac{b^2+c^2-a^2}{2bc}$ $\Big) + 2ac \Big($ $\frac{a^2+c^2-b^2}{2ac}$ $\Big) + 2ab \Big($ $\frac{a^2+b^2-c^2}{2ab}$ $\Big)$

$= ( b^2+c^2-a^2 + a^2+c^2-b^2 + a^2+b^2-c^2)$

$= a^2 + b^2 + c^2$

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Question 8: $(c^2 - a^2 + b^2) \tan A = ( a^2 - b^2 +c^2) \tan B = (b^2 -c^2+a^2) \tan C$

$(c^2 - a^2 + b^2) \tan A = (c^2 - a^2 + b^2)$ $\frac{\sin A}{\cos A}$ $= (c^2 - a^2 + b^2) \Big[$ $\frac{ak (2bc)}{(c^2 - a^2 + b^2)}$ $\Big] = 2abck$

$( a^2 - b^2 +c^2) \tan B = ( a^2 - b^2 +c^2)$ $\frac{\sin B}{\cos B}$ $= ( a^2 - b^2 +c^2) \Big[$ $\frac{bk (2ca)}{( a^2 - b^2 +c^2)}$ $\Big] = 2abck$

$(b^2 -c^2+a^2) \tan C = (b^2 -c^2+a^2)$ $\frac{\sin C}{\cos C}$ $= (b^2 -c^2+a^2) \Big[$ $\frac{ck (2ab)}{(b^2 -c^2+a^2)}$ $\Big] = 2abck$

Hence proved.

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Question 9: $\frac{c - b \cos A}{b - c \cos A}$ $=$ $\frac{\cos B}{\cos C}$

LHS $=$ $\frac{c - b \cos A}{b - c \cos A}$

$=$ $\frac{k \sin C - k \sin B \cos A}{k \sin B - k \sin C \cos A}$

$=$ $\frac{ \sin C - \sin B \cos A}{ \sin B - \sin C \cos A}$

$=$ $\frac{ \sin (\pi - ( A+B)) - \sin B \cos A}{ \sin (\pi - ( A +C)) - \sin C \cos A}$

$=$ $\frac{ \sin ( A+B) - \sin B \cos A}{ \sin ( A +C) - \sin C \cos A}$

$=$ $\frac{ \sin A \cos B + \cos A \sin B - \sin B \cos A}{ \sin A \cos C + \cos A \sin C - \sin C \cos A}$

$=$ $\frac{\sin A \cos B}{\sin A \cos C}$

$=$ $\frac{\cos B}{\cos C}$ $=$ RHS. Hence proved.

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Question 10: $a( \cos B + \cos C - 1) + b( \cos C + \cos A - 1) + c( \cos C + \cos A - 1) = 0$

In $\triangle ABC$, we have

$a = b \cos C + c \cos B,$     $b = c \cos A + a \cos C,$     $c = a \cos B + b \cos A$

LHS $= a( \cos B + \cos C - 1) + b( \cos C + \cos A - 1) + c( \cos C + \cos A - 1)$

$= a \cos B + a \cos C - a + b \cos C + b \cos A - b + c \cos C + c\cos A - c$

$= (b \cos C + c \cos B) + ( c \cos A + a \cos C) + ( a \cos B + b \cos A) - (a + b + c)$

$= (a + b + c) - (a + b + c) = 0 =$ RHS. Hence proved.

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Question 11: $a \cos A + b \cos B + c \cos C = 2b \sin A \sin C$

LHS $= a \cos A + b \cos B + c \cos C$

$= k \sin A \cos A + k \sin B \cos B + k \sin C \cos C$

$=$ $\frac{k}{2}$ $\Big[ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ \sin 2A + \sin 2B + 2\sin C \cos C \Big]$

$=$ $\frac{k}{2}$ $\Big[ 2 \sin (A+B) \cos (A-B) + 2\sin C \cos C \Big]$

$= k \Big[ \sin (\pi - C) \cos (A-B) + \sin C \cos (\pi - (A+B)) \Big]$

$= k \Big[ \sin C \cos (A-B) + \sin C \cos (A+B) \Big]$

$= k \sin C \Big[ \cos (A-B) + \cos (A+B) \Big]$

$= k \sin C 2\sin A \sin B$

$= 2 ( k \sin A) \sin B \sin C$

$= 2 a \sin B \sin C =$ RHS. Hence proved.

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Question 12: $a^2 = (b^2 + c^2)- 4 bc \cos^2$ $\frac{A}{2}$

We know $\cos A =$ $\frac{b^2 + c^2 - a^2}{2bc}$

$\Rightarrow 2 bc \cos A = b^2 + c^2 - a^2$

$\Rightarrow a^2 = b^2 + c^2 -2 bc \cos A$

$\Rightarrow a^2 = b^2 + c^2 -2bc( 2 \cos^2$ $\frac{A}{2}$ $-1)$

$\Rightarrow a^2 =b^2 + c^2 +2bc - 4 bc \cos^2$ $\frac{A}{2}$

$\Rightarrow a^2 =(b+c)^2 - 4 bc \cos^2$ $\frac{A}{2}$

Hence proved.

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Question 13: $4 \Big( bc \ \cos^2$ $\frac{A}{2}$ $+ ca\ \cos^2$ $\frac{B}{2}$ $+ ab\ \cos^2$ $\frac{C}{2}$ $\Big) = (a+ b+c)^2$

LHS $= 4 \Big[ bc \ \cos^2$ $\frac{A}{2}$ $+ ca\ \cos^2$ $\frac{B}{2}$ $+ ab\ \cos^2$ $\frac{C}{2}$ $\Big]$

$= 2 \Big[ bc \ 2\cos^2$ $\frac{A}{2}$ $+ ca\ 2\cos^2$ $\frac{B}{2}$ $+ ab\ 2\cos^2$ $\frac{C}{2}$ $\Big]$

$= 2 \Big[ bc \ (1+\cos A) + ca\ (1+\cos B) + ab\ (1+\cos C) \Big]$

$= 2bc + 2bc \cos A + 2ca + 2ca \cos B + 2ab+ 2ab\cos C$

$= 2bc + 2bc \Big[$ $\frac{b^2 + c^2 - a^2}{2bc}$ $\Big] + 2ca + 2ca \Big[$ $\frac{c^2 + a^2 - b^2}{2ca}$ $\Big] + 2ab+ 2ab \Big[$ $\frac{a^2 + b^2 - c^2}{2ab}$ $\Big]$

$= 2bc + b^2 + c^2 - a^2 + 2ca + c^2 + a^2 - b^2 + 2ab+ a^2 + b^2 - c^2$

$= 2bc + 2ac + 2 ab + b^2 + c^2 + a^2$

$= (a+b+c)^2 =$ RHS. Hence proved.

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Question 14:  In $\triangle ABC$ prove that

$\sin^3 A \cos ( B-C) + \sin^3 B \cos ( C-A) + \sin^3 C \cos ( A-B) = 3 \sin A \sin B \sin C$

LHS $= \sin^3 A \cos ( B-C) + \sin^3 B \cos ( C-A) + \sin^3 C \cos ( A-B)$

$= \sin^2 A \sin ( \pi - (B+C)) \cos (B-C) + \sin^2 B \sin ( \pi - (C+A)) \cos (C-A) + \sin^2 C \sin ( \pi - (A+B)) \cos (A-B)$

$= \sin^2 A \sin (B+C) \cos (B-C) + \sin^2 B \sin (C+A) \cos (C-A) + \sin^2 C \sin (A+B) \cos (A-B)$

$= \frac{1}{2} \Big[ \sin^2 A ( \sin 2B + \sin 2C ) + \sin^2 B ( \sin 2C + \sin 2A ) + \sin^2 C ( \sin 2A + \sin 2B ) \Big]$

$= \frac{1}{2} \Big[ \sin^2 A ( 2 \sin B \cos B + 2\sin C \cos C) + \sin^2 B ( 2 \sin C \cos C + 2\sin A \cos A ) + \sin^2 C ( 2 \sin A \cos A + 2\sin B \cos B ) \Big]$

$= \frac{1}{2} \Big[ 2k^3a^2b \cos B + 2k^3a^2c \cos C + 2k^3b^2c \cos C + 2k^3b^2a \cos A + 2k^3c^2a \cos A + 2k^3c^2b \cos B \Big]$

$= \frac{1}{2} \Big[ 2k^3ab ( a \cos B + b \cos A) + 2k^3ac ( a \cos C + c \cos A) + 2k^3bc ( b \cos C + c \cos B) \Big]$

$= k^3 abc + k^3 abc + k^3 abc$

$= 3 ( ka) ( kb) ( kc)$

$= 3 \sin A \sin B \sin C =$ RHS. Hence proved.

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Question 15: In any $\triangle ABC$, $\frac{b+c}{12}$ $=$ $\frac{c+a}{13}$ $=$ $\frac{a+b}{15}$, then prove that $\frac{\cos A}{2}$ $=$ $\frac{\cos B}{7}$ $=$ $\frac{\cos C}{11}$

Let $\frac{b+c}{12}$ $=$ $\frac{c+a}{13}$ $=$ $\frac{a+b}{15}$ $= k$

$\Rightarrow b+c = 12k, c+a = 13k, a+b = 15k$

$b+c + c+ a + a + b = 12k + 13k + 15k$

$\Rightarrow 2 ( a + b + c) = 40 k$

$\Rightarrow a + b + c = 20 k$

$\therefore a + 12k = 20 k \Rightarrow a = 8k$

$b + 13k = 20 k \Rightarrow b = 7k$

$c + 15k = 20 k \Rightarrow c = 5 k$

$\therefore$  $\cos A =$ $\frac{b^2 + c^2 - a^2}{2bc}$ $=$ $\frac{49k^2 + 25k^2 - 64k^2}{2 \times 7k \times 5k}$ $=$ $\frac{1}{7}$

$\cos B =$ $\frac{c^2 + a^2 - b^2}{2ca}$ $=$ $\frac{64k^2 + 25k^2 - 49k^2}{2 \times 8k \times 5k}$ $=$ $\frac{1}{2}$

$\cos C =$ $\frac{a^2 + b^2 - c^2}{2ab}$ $=$ $\frac{49k^2 + 64k^2 - 25k^2}{2 \times 8k \times 7k}$ $=$ $\frac{11}{14}$

$\therefore \cos A : \cos B : \cos C =$ $\frac{1}{7}$ $:$ $\frac{1}{2}$ $:$ $\frac{11}{14}$ or

$\cos A : \cos B : \cos C = 2:7:11$

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Question 16: In any $\triangle ABC$, if $\angle B = 60^o$, prove that $(a+b+c)(a-b+c) = 3ca$

Given $\angle B = 60^o$

$\cos B =$ $\frac{1}{2}$ $\Rightarrow$ $\frac{c^2 + a^2 - b^2}{2ca}$ $=$ $\frac{1}{2}$

$\Rightarrow a^2 + c^2 - b^2 = ac$

$\Rightarrow a^2 + c^2 -ac = b^2$

Now $(a+b+c)(a-b+c) = 2ac$

$\Rightarrow a^2 + ab + ca - ab - b^2 -bc + ca + bc + c^2 = 3ac$

$\Rightarrow (a^2 + c^2 -ac) - b^2 = 0$

$\Rightarrow b^2 - b^2 = 0$

Hence proved.

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Question 17: In any $\triangle ABC$, $\cos^2 A + \cos^2 B + \cos^2 C = 1$, prove that the triangle is right angled.

$\cos^2 A + \cos^2 B + \cos^2 C = 1$

$\Rightarrow \cos^2 A + \cos^2 B - (1 - \cos^2 C) = 0$

$\Rightarrow \cos^2 A + \cos^2 B - \sin^2 C = 0$

$\Rightarrow \cos^2 A + \cos ( B+C) \cos ( B - C) = 0$

$\Rightarrow \cos^2 A + \cos ( \pi - A) \cos ( B - C) = 0$

$\Rightarrow \cos^2 A - \cos A \cos ( B - C) = 0$

$\Rightarrow \cos A \Big[ \cos A - \cos ( B - C) \Big] = 0$

$\Rightarrow \cos A \Big[ \cos (\pi - ( B+C)) - \cos ( B - C) \Big] = 0$

$\Rightarrow \cos A \Big[ -\cos ( B+C) - \cos ( B - C) \Big] = 0$

$\Rightarrow -2\cos A \Big[ 2 \cos B \cos C \Big] = 0$

$\Rightarrow \cos A \cos B \cos C = 0$

This means either $\cos A = 0 \ or \ \cos B = 0 \ or \ \cos C = 0$

Hence one of the angles is $90^o$. Therefore the triangle is a right angled triangle.

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Question 18: In any $\triangle ABC$, if $\cos C =$ $\frac{\sin A}{2 \sin B}$, prove that the triangle is isosceles.

Using sine rule,

$\frac{\sin A}{a}$ $=$ $\frac{\sin B}{b}$ $=$ $\frac{\sin C}{c}$ $= k$

$\Rightarrow \sin A = ak, \sin B = bk, \sin C = ck$

Now $\cos C =$ $\frac{\sin A}{2 \sin B}$

$2 \sin B \cos C = \sin A$

$2 (kb) \Big($ $\frac{a^2 + b^2 - c^2}{2ab}$ $\Big) = ka$

$\Rightarrow a^2 + b^2 -c^2 = a^2$

$\Rightarrow b^2 = c^2$

$\Rightarrow b = c$

$\therefore$ the triangle is Isosceles.

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Question 19: Two ships leave a port at the same time. One goes $24$ km/hr in the direction of $N 38^o E$ and the other travels $32$ km/hr in the direction of $S 52^o E$. Find the distance between the ships at the end of $3$ hrs.

Let $P$ and $Q$ be the position of two ships at the end of $3$ hours.

$OP = 3 \times 24 = 72$ km

$OQ = 3 \times 32 = 96$ km

Using cosine formula in $\triangle OPQ$

$PQ^2 = OP^2 + OQ^2 - 2OP \times OQ \cos 90^o$

$\Rightarrow PQ^2 = 72^2 + 96^2 - 2 \times 72 \times 96 \times 1$

$\Rightarrow PQ^2 = 14400$

$\Rightarrow PQ = 120$ km