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.

Answer:

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.

Answer:

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 .

Answer:

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 .

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

(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}

Answer:

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

Answer:

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

Answer:

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}

Answer:

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

Answer:

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

Answer:

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}

Answer:

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

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

Adding all the three

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

Answer:

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.

Answer:

\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.

Answer:

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.

Answer:

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

OP = 3 \times 24 = 72 km2020-05-20_21-55-00

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