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

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

Answer:

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

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

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

Answer:

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

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

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

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

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

Answer:

$\displaystyle \text{Given } a = 4, b = 6 \text{ and } c = 8$

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

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

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

$\displaystyle \text{LHS } = 8 \cos A + 16 \cos B + 4 \cos C$

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

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

Answer:

$\displaystyle \text{Given } a = 18, b = 24 \text{ and } c = 30$

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

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

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

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

Answer:

$\displaystyle \text{RHS } = c^2 - a^2$

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

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

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

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

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

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

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

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

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

Answer:

$\displaystyle \text{LHS } = c ( a \cos B - b \cos A)$

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

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

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

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

$\displaystyle = a^2 - b^2 = \text{ RHS. Hence proved. }$

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

Answer:

$\displaystyle \text{LHS } = 2( bc \cos A + ca \cos B + ab \cos C)$

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

Hence proved.

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

Answer:

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

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

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

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

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

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

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

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

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

Answer:

In $\displaystyle \triangle ABC$, we have

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

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

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

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

$\displaystyle = (a + b + c) - (a + b + c) = 0 = \text{ RHS. Hence proved. }$

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

Answer:

$\displaystyle \text{LHS } = a \cos A + b \cos B + c \cos C$

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

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

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

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

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

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

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

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

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

$\displaystyle = 2 a \sin B \sin C = \text{ RHS. Hence proved. }$

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

Answer:

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

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

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

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

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

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

Hence proved.

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

Answer:

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

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

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

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

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

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

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

$\displaystyle = (a+b+c)^2 = \text{ RHS. Hence proved. }$

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Question 14: In $\displaystyle \triangle ABC$ prove that

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

Answer:

$\displaystyle \text{LHS } = \sin^3 A \cos ( B-C) + \sin^3 B \cos ( C-A) + \sin^3 C \cos ( A-B)$

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

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

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

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

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

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

$\displaystyle = k^3 abc + k^3 abc + k^3 abc$

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

$\displaystyle = 3 \sin A \sin B \sin C = \text{ RHS. Hence proved. }$

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

Answer:

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

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

Adding all the three

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

$\displaystyle \text{Given } \angle B = 60^o$

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

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

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

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

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

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

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

Hence proved.

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

Using sine rule,

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

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

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

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

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

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

$\displaystyle \Rightarrow b^2 = c^2$

$\displaystyle \Rightarrow b = c$

$\displaystyle \therefore$ the triangle is Isosceles.

$\displaystyle \\$

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

Answer:

Let $\displaystyle P \text{ and } Q$ be the position of two ships at the end of $\displaystyle 3$ hours.

$\displaystyle OP = 3 \times 24 = 72 \text{ km }$

$\displaystyle OQ = 3 \times 32 = 96 \text{ km }$

Using cosine formula in $\displaystyle \triangle OPQ$

$\displaystyle PQ^2 = OP^2 + OQ^2 - 2OP \times OQ \cos 90^{\circ}$

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

$\displaystyle \Rightarrow PQ^2 = 14400$

$\displaystyle \Rightarrow PQ = 120 \text{ km }$