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Class 11 RD Sharma Solutions – Chapter 10 Sine and Cosine Formulae and Their Applications – Exercise 10.1 | Set 2

  • Last Updated : 16 May, 2021

Question 16: In △ABC, prove the following:

a2 (cos2 B – cos2 C) + b(cos2 C – cos2 A) + c(cos2 A – cos2 B) = 0

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

= a2 (cos2 B – cos2 C) + b(cos2 C – cos2 A) + c(cos2 A – cos2 B)



By using trigonometric formula,

cos2 a = 1 – sin2 a

= λ2 sin2 A(1-sin2 B – (1-sin2 C)) + λ2 sin2 B(1-sin2 C – (1-sin2 A)) + λ2 sin2 C(1-sin2 A – (1-sin2 B))

= λ2 sin2 A(1-sin2 B – 1+ sin2 C) + λ2 sin2 B(1-sin2 C – 1 + sin2 A) + λ2 sin2 C(1-sin2 A – 1+sin2 B)

= λ2 sin2 A(sin2 C – sin2 B) + λ2 sin2 B(sin2 A – sin2 C) + λ2 sin2 C(sin2 B – sin2 A)

= λ2 (sin2 A sin2 C – sin2 A sin2 B + sin2 B sin2 A – sin2 B sin2 C + sin2 C sin2 B – sin2 C sin2 A)

= λ2 (0)

=0



As LHS = RHS

Hence, proved!!

Question 17: In △ABC, prove the following:

b cos B + c cos C = a cos (B-C)

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

= b cos B + c cos C

= λ sin B cos B + λ sin C cos C

= λ (sin B cos B + sin C cos C)

=\frac{\lambda}{2}  (2 sin B cos B + 2 sin C cos C)



By using trigonometric formula,

2 sin a cos a = sin 2a

=\frac{\lambda}{2}  (sin 2B + sin 2C) ………………………..(1)

Now considering RHS, we have

= a cos (B-C)

= λ sin A cos (B-C)

=\frac{\lambda}{2}  (2 sin A cos (B-C))

By using trigonometric formula,

2 sin a cos b = sin (a+b) + sin(a-b)

= \frac{\lambda}{2}(sin (A+B-C) + sin(A-B+C))\\ = \frac{\lambda}{2}(sin ((\pi-C)-C) + sin((\pi-B)-B))\\ = \frac{\lambda}{2}(sin (\pi-C-C) + sin(\pi-B-B))\\ = \frac{\lambda}{2}(sin (\pi-2C) + sin(\pi-2B))



=\frac{\lambda}{2}  (sin 2C + sin2B) ……………………….(2)

As LHS = RHS

Hence, proved!!

Question 18: In △ABC, prove the following:

\frac{cos(2A)}{a^2} - \frac{cos(2B)}{b^2} = \frac{1}{a^2} - \frac{1}{b^2}

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

= \frac{cos(2A)}{a^2} - \frac{cos(2B)}{b^2}

By using trigonometric formula,

cos 2a = 1-2sin2a



= \frac{1-2sin^2A}{a^2} - \frac{1-2sin^2B}{b^2}\\ = \frac{1-2(\frac{a}{\lambda})^2}{a^2} - \frac{1-2(\frac{b}{\lambda})^2}{b^2}\\ = \frac{\frac{\lambda^2-2a^2}{\lambda^2}}{a^2} - \frac{\frac{\lambda^2-2b^2}{\lambda^2}}{b^2}\\ = \frac{1}{\lambda^2}(\frac{\lambda^2-2a^2}{a^2} - \frac{\lambda^2-2b^2}{b^2})\\ = \frac{1}{\lambda^2}(\frac{\lambda^2b^2-2a^2b^2 - a^2\lambda^2+2a^2b^2}{a^2b^2})\\ = \frac{1}{\lambda^2}(\frac{\lambda^2b^2- a^2\lambda^2}{a^2b^2})\\ = \frac{b^2- a^2}{a^2b^2}\\ = \frac{1}{a^2} - \frac{1}{b^2}

As LHS = RHS

Hence, proved!!

Question 19: In △ABC, prove the following:

\frac{cos^2B-cos^2C}{b+c} + \frac{cos^2C-cos^2A}{c+a} + \frac{cos^2A-cos^2B}{a+b} = 0

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS, we have

= \frac{cos^2B-cos^2C}{b+c} + \frac{cos^2C-cos^2A}{c+a} + \frac{cos^2A-cos^2B}{a+b}

Now taking,

\frac{cos^2B-cos^2C}{b+c} = \frac{cos^2B-cos^2C}{\lambda(sin B + sin C)}\\ = \frac{cos^2B-cos^2C}{\lambda(sin B + sin C)}\\ = \frac{(cosB-cosC)(cosB+cosC)}{\lambda(sin B + sin C)}



By using trigonometric formula,

cos a + cos b = 2 cos(\frac{a+b}{2})  cos(\frac{a-b}{2})

cos a – cos b = -2 sin(\frac{a+b}{2})  sin(\frac{a-b}{2})

sin a + sin b = 2 sin(\frac{a+b}{2})  cos(\frac{a-b}{2})

= \frac{(2 cos (\frac{B+C}{2}) cos(\frac{B-C}{2}))(-2 sin (\frac{B+C}{2}) sin(\frac{B-C}{2}))}{\lambda(2 sin (\frac{B+C}{2}) cos(\frac{B-C}{2}))}\\ = \frac{- 2 cos (\frac{B+C}{2}) sin (\frac{B-C}{2}}{\lambda}

By using trigonometric formula,

2 cos(\frac{a+b}{2})  sin(\frac{a-b}{2})  = sin a – sin b

= \frac{- (sin B - sin C)}{\lambda}

= \frac{sin C - sin B}{\lambda}  ………………(1)

Similarly, we can prove,



\frac{cos^2C-cos^2A}{c+a} = \frac{sin A - sin C}{\lambda}  ……………….(2)

\frac{cos^2A-cos^2B}{a+b} = \frac{sin B - sin A}{\lambda}  ………………..(3)

Adding (1), (2) and (3), we get

\frac{sin C - sin B}{\lambda} + \frac{sin A - sin C}{\lambda} + \frac{sin B - sin A}{\lambda}

= 0

As LHS = RHS

Hence, proved!!

Question 20: In △ABC, prove the following:

a sin \frac{A}{2} sin \frac{B-C}{2} + b sin\frac{B}{2} sin \frac{C-A}{2} + c sin\frac{C}{2} sin \frac{A-B}{2} = 0

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda



Considering LHS, we have

= a sin \frac{A}{2} sin \frac{B-C}{2} + b sin\frac{B}{2} sin \frac{C-A}{2} + c sin\frac{C}{2} sin \frac{A-B}{2}

Now taking,

a sin \frac{A}{2} sin \frac{B-C}{2} = a sin \frac{\pi-(B+C)}{2} sin \frac{B-C}{2}\\ = a sin (\frac{\pi}{2}-\frac{(B+C)}{2}) sin \frac{B-C}{2}\\ = a cos (\frac{B+C}{2}) sin \frac{B-C}{2}

=\frac{2}{2}  a cos(\frac{B+C}{2})  sin(\frac{B-C}{2})

=\frac{1}{2}  (sin B – sin C)

=\frac{1}{2}  (a sin B – a sin C)

=\frac{1}{2}  (a sin B – a sin C)

=\frac{1}{2}  (b sin A – a sin C) ………………..(1)

Similarly, we can prove,



bsin\frac{B}{2}  sin(\frac{C-A}{2})  =\frac{1}{2}  (b sin C – b sin A) ……………….(2)

csin\frac{C}{2}  sin(\frac{A-B}{2})  =\frac{1}{2}  (a sin C – b sin C) ………………..(3)

Adding (1), (2) and (3), we get

\frac{1}{2}  (b sin A – a sin C) +\frac{1}{2}  (b sin C – b sin A) +\frac{1}{2}  (a sin C – b sin C)

=\frac{1}{2}  (b sin A – a sin C + b sin C – b sin A + a sin C – b sin C)

= 0

As LHS = RHS

Hence, proved!!

Question 21: In △ABC, prove the following:

\frac{b secB+c sec C}{tan B +tan C} = \frac{c sec C+ a sec A}{tan C +tan A} = \frac{a secA+b sec B}{tan A +tan B}

Solution:

According to the sine rule



\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering equation, we have

\frac{b secB+c sec C}{tan B +tan C} = \frac{c sec C+ a sec A}{tan C +tan A} = \frac{a secA+b sec B}{tan A +tan B}

Now taking,

\frac{b secB+c sec C}{tan B +tan C} = \frac{\lambda sin B secB+ \lambda sin C sec C}{tan B +tan C}\\ = \frac{\lambda sin B \frac{1}{cos B} + \lambda sin C \frac{1}{cos C}}{tan B +tan C}\\ = \frac{\lambda tan B+ \lambda tan C}{tan B +tan C}\\ = \frac{\lambda (tan B+tan C)}{tan B +tan C}\\ = \lambda

Similarly, we can prove,

\frac{c sec C+ a sec A}{tan C +tan A} = \lambda ...................(2)\\ \frac{a secA+b sec B}{tan A +tan B} = \lambda ....................(3)

From, (1), (2) and (3), we get

\frac{b secB+c sec C}{tan B +tan C} = \frac{c sec C+ a sec A}{tan C +tan A} = \frac{a secA+b sec B}{tan A +tan B}

Hence, proved!!



Question 22: In △ABC, prove the following:

a cos A + b cos B + c cos C = 2b sin A sin C = 2c sin A sin B

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering LHS equation, we have

a cos A + b cos B + c cos C = λ sin A cos A + λ sin B cos B + λ sin C cos C

= λ (sin A cos A + sin B cos B + sin C cos C)

=\frac{2\lambda}{2}  (sin A cos A + sin B cos B + sin C cos C)

=\frac{\lambda}{2}  (2 sin A cos A + 2 sin B cos B + 2 sin C cos C)

By using trigonometric formula,

2 sin a cos a = sin 2a

=\frac{\lambda}{2}  (sin 2A + sin 2B + 2 sin C cos C)

By using trigonometric formula,

sin a + sin b = 2 sin(\frac{a+b}{2})  cos(\frac{a-b}{2})

=\frac{\lambda}{2} (2 sin (\frac{2A+2B}{2}) cos(\frac{2A-2B}{2})  + 2 sin C cos C)

=\frac{\lambda}{2}  (2 sin (A+B) cos(A-B) + 2 sin C cos C)

=\frac{\lambda}{2}  (2 sin (π-C) cos(A-B) + 2 sin C cos C)

=\frac{\lambda}{2}  (2 sin C cos(A-B) + 2 sin C cos C)

=\frac{2\lambda sinC}{2}  (cos(A-B) + cos C)

= λ sin C (cos(A-B) + cos (π-(A+B)))

= λ sin C (cos(A-B) + (-cos (A+B)))



= λ sin C (cos(A-B) – cos (A+B))

= λ sin C (2 sin A sin B)

= 2 λ sin A sin B sin C

Now putting λ sin C = c and λ sin B = b, we get

2 c sin A sin B and 2 b sin A sin C

Hence, proved!!

Question 23: In △ABC, prove the following:

a (cos B cos C + cos A) = b (cos C cos A + cos B) = c (cos C cos A + cos C)

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering equation, we have

a (cos B cos C + cos A) = λ sin A (cos B cos C + cos A)

= λ (sin A cos B cos C + sin A cos A)

= λ (\frac{cos C}{2}  (2 sin A cos B) + sin A cos A)

= λ (\frac{cos C}{2}  (sin (A+B) + sin (A-B)) + sin A cos A)

= λ (\frac{1}{2}  (cos C sin (A+B) + cos C sin (A-B)) + sin A cos A)

= λ (\frac{1}{2}  (\frac{1}{2} (2 cos C sin (A+B) + 2 cos C sin (A-B))) + sin A cos A)

= λ (\frac{1}{2}  (2 cos C sin (A+B) + 2 cos C sin (A-B)) + sin A cos A)

By using trigonometric formula,

2 sin a cos b = sin (a+b) + sin (a-b)

= λ (\frac{1}{4}  (sin (A+B+C) + sin (A+B-C) + sin (A-B+C) + sin (A-B-C)) + sin A cos A)

= λ (\frac{1}{4}  (sin (π) + sin ((π-C)-C) + sin ((π-B)-B) + sin (A-(B+C)) +\frac{sin 2A}{2}  )

= λ (\frac{1}{4}  (0 + sin (π-2C) + sin (π-2B) + sin (2A-π) +\frac{sin 2A}{2}  )

= λ (\frac{1}{4}  (sin 2C + sin 2B – sin 2A +\frac{sin 2A}{2}  )

=\frac{\lambda}{4}  (sin 2C + sin 2B + sin 2A)

Similarly,

b (cos C cos A + cos B) =\frac{\lambda}{4}  (sin 2C + sin 2B + sin 2A)

c (cos C cos A + cos C) =\frac{\lambda}{4}  (sin 2C + sin 2B + sin 2A)

Hence, proved!!

Question 24: In △ABC, prove the following:

a (cos C – cos B) = 2 (b-c)cos^2(\frac{A}{2})

Solution:

According to the sine rule



\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering equation, we have

a (cos C – cos B) = λ sin A (cos C – cos B)

= λ (sin A cos C – sin A cos B)

=\frac{\lambda}{2}  (2 sin A cos C – 2 sin A cos B)

By using trigonometric formula,

2 sin a cos b = sin (a+b) + sin (a-b)

=\frac{\lambda}{2}  (sin (A+C) + sin (A-C) – (sin (A+B) + sin (A-B)))

=\frac{\lambda}{2}  (sin (A+C) + sin (A-C) – sin (A+B) – sin (A-B))

=\frac{\lambda}{2}  (sin (π-B) + sin (A-C) – sin (π-C) – sin (A-B))

=\frac{\lambda}{2}  (sin B – sin C + sin (A-C) – sin (A-B))

By using trigonometric formula,

sin a – sin b = 2 sin(\frac{a-b}{2})  cos(\frac{a+b}{2})

= \frac{\lambda}{2} (2 sin (\frac{B-C}{2}) cos(\frac{B+C}{2}) + 2 sin (\frac{A-C-(A-B)}{2}) cos(\frac{A-C+A-B}{2}))\\ = \frac{\lambda}{2} (2 sin (\frac{B-C}{2}) cos(\frac{B+C}{2}) + 2 sin (\frac{B-C)}{2}) cos(\frac{2A-(B+C)}{2}))

= λsin (\frac{B-C}{2}) (cos(\frac{\pi-A}{2}) + cos(\frac{\pi-3A}{2}))

= λsin (\frac{B-C}{2}) (cos(\frac{π}{2}-\frac{A}{2}) + cos(\frac{π}{2}-\frac{3A}{2}))

= λ sin(\frac{B-C}{2}) (sin(\frac{A}{2}) + sin(\frac{3A}{2}))

By using trigonometric formula,

sin a + sin b = 2 sin(\frac{a+b}{2})  cos(\frac{a-b}{2})

= λ sin(\frac{B-C}{2})
(2 sin (\frac{\frac{3A}{2}+\frac{A}{2}}{2}) cos(\frac{\frac{3A}{2}-\frac{A}{2}}{2}))



= λ sin(\frac{B-C}{2})
(2 sin (\frac{\frac{4A}{2}}{2}) cos(\frac{\frac{2A}{2}}{2}))

= λ sin(\frac{B-C}{2})  (2 sin A cos(\frac{A}{2})  )

= 2 λ sin(\frac{B-C}{2})  (2 sin(\frac{A}{2})  cos(\frac{A}{2})  ) cos(\frac{A}{2})

= 4 λ sin(\frac{B-C}{2})  sin(\frac{A}{2})  [Tex]cos^2(\frac{A}{2})[/Tex]

= 4 λ sin(\frac{B-C}{2})  cos(\frac{\pi}{2}-\frac{A}{2})  [Tex]cos^2(\frac{A}{2})[/Tex]

= 4 λ sin(\frac{B-C}{2})  cos(\frac{\pi-A}{2})  [Tex]cos^2(\frac{A}{2})[/Tex]

= 4 λ sin(\frac{B-C}{2})  cos(\frac{B+C}{2})  [Tex]cos^2(\frac{A}{2})[/Tex]

= 2 λ (2 sin(\frac{B-C}{2})  cos(\frac{B+C}{2})  )cos^2(\frac{A}{2})

By using trigonometric formula,

2 cos(\frac{a+b}{2})  sin(\frac{a-b}{2})  = sin a – sin b



= 2 λ (sin B – sin A)cos^2(\frac{A}{2})

= 2 (λ sin B – λ sin A)cos^2(\frac{A}{2})

= 2 (b – a)cos^2(\frac{A}{2})

As LHS = RHS

Hence, proved!!

Question 25: In △ABC, prove the following:

b cos θ = c cos(A-θ)+a cos(C+θ)

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering RHS, equation, we have

c cos(A-θ)+a cos(C+θ) = λ sin C cos(A-θ) + λ sin A cos(C+θ)

= λ (sin C cos(A-θ) + sin A cos(C+θ))

=\frac{\lambda}{2} (2 sin C cos(A-θ) + 2 sin A cos(C+θ))

By using trigonometric formula,

2 sin a cos b = sin (a+b) + sin (a-b)

= \frac{\lambda}{2} (sin (C+A-θ) + sin (C-(A-θ)) + sin (A+C+θ) + sin (A-(C-θ)))\\ = \frac{\lambda}{2} (sin (C+A-θ) + sin (C-A+θ) + sin (A+C+θ) + sin (A-C+θ))\\ = \frac{\lambda}{2} (sin (π-B-θ) + sin (C+θ-A) + sin (π-B+θ) - sin (C+θ-A))\\ = \frac{\lambda}{2} (sin (π-(B+θ)) + sin (π-(B-θ)))\\ = \frac{\lambda}{2} (sin (B+θ) + sin (B-θ))

By using trigonometric formula,

sin (a+b) + sin (a-b) = 2 sin a cos b

=\frac{\lambda}{2} (2 sin B cos θ)

= λ sin B cos θ

= b cos θ

As LHS = RHS

Hence, proved!!

Question 26: In △ABC, if sin2A + sin2B = sin2C. Show that the triangle is right-angled.

Solution:

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

Considering equation, we have

sin2 A + sin2 B = sin2 C

(\frac{a}{\lambda})^2 + (\frac{b}{\lambda})^2 = (\frac{c}{\lambda})^2

a2 + b2 = c2

Hence, proved, the triangle is right-angled as c as hypotenuse.



Question 27: In △ABC, if a2, b2, and c2 are in AP. Prove that cot A, cot B, and cot C are also in AP.

Solution:

We have a2, b2 and c2 in AP

2a2, 2b2 and 2c2 are also in AP

(a2+b2+c2)-2a2, (a2+b2+c2)-2b2 and (a2+b2+c2)-2c2 are also in AP

b2+c2-a2, a2+c2-b2 and a2+b2-c2 are also in AP

\frac{b^2+c^2-a^2}{2abc}  ,\frac{a^2+c^2-b^2}{2abc}  and\frac{a^2+b^2-c^2}{2abc}  are also in AP

\frac{1}{a}(\frac{b^2+c^2-a^2}{2bc})  ,\frac{1}{b}(\frac{a^2+c^2-b^2}{2ac})  and\frac{1}{c}(\frac{a^2+b^2-c^2}{2ab})  are also in AP

According to the cosine rule

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

\frac{1}{a}(cos A)  ,\frac{1}{b}(cos B)  and\frac{1}{c}(cos C)  are also in AP

\frac{\lambda}{a}(cos A)  ,\frac{\lambda}{a}(cos B)  and\frac{\lambda}{a}(cos C)  are also in AP

According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

\frac{cos A}{sin A}  ,\frac{cos B}{sin B}  and\frac{cos C}{sin C}  are also in AP

cot A, cot B and cot C are also in AP

Hence, proved !!

Question 28: The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Solution:

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.

The total height of the tree is x+y.



In △ABC, ∠C = 30° and ∠B = 90°

∠A = 60° (Triangle angle sum property)

According to the sine rule

\frac{AB}{sin \hspace{0.1cm}30\degree} = \frac{BC}{sin \hspace{0.1cm}60\degree} = \frac{AC}{sin \hspace{0.1cm}90\degree}\\ \frac{x}{\frac{1}{2}} = \frac{15}{\frac{\sqrt{3}}{2}} = \frac{y}{1}

2x =\frac{30}{\sqrt{3}}  = y

Hence, x =\frac{30}{2\sqrt{3}} = \frac{15}{\sqrt{3}} = \frac{15}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} = \frac{15\sqrt{3}}{3}  = 5√3

and y =\frac{30}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3}  = 10√3

So, the height of tree x+y = 5√3+10√3

= 15√3 m

Question 29: At the foot of a mountain, the elevation of it summit is 45°, after ascending 1000m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

Solution:

Suppose, AB is a mountain of height t+x.

c = 1000m

In △DFC,

sin 30° =\frac{x}{1000}

\frac{1}{2} = \frac{x}{1000}

x =\frac{1000}{2}  = 500 m

And, tan 30° =\frac{x}{y}

\frac{1}{\sqrt{3}} = \frac{500}{y}

y = 500√3 m

Now, In △ABC

tan 45° =\frac{t+x}{y+z}

1 =\frac{t+500}{y+z}

500√3+z = t+500

500(√3-1)+z = t …………………..(1)

Now, In △ADE

tan 60° =\frac{t}{z}

√3 =\frac{t}{z}

t = √3z ………………….(2)

From (1) and (2), we get

z = 500 m

t = 500√3 m

So, the height of the mountain = t+x = 500√3 + 500 = 500(√3+1)m

Question 30: A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be γ. Show that the height of the peak above the ground is\frac{c \hspace{0.1cm}sin \alpha \hspace{0.1cm}sin(\gamma-\beta)}{sin\gamma - \alpha}

Solution:

Suppose, AB is a peak whose height above the ground is t+x,

In △DFC,

sin β =\frac{x}{c}

x = c sin β ………………………..(1)

And, tan β =\frac{x}{y}



y = \frac{x}{tan β} = \frac{c sin β}{tan β}  = c cos β ………………………..(2)

Now, In △ADE

tan γ =\frac{t}{z}

z = t cot γ ………………………(3)

Now, In △ABC

tan α=\frac{t+x}{y+z}

t +x = tan α (y+z)

From (1), (2) and (3), we get

t + c sin β = tan α (c cos β+t cot γ)

t + c sin β = c tan α cos β + t tan α cot γ

t – t tan α cot γ = c tan α cos β – c sin β

t(1 – tan α cot γ) = c (tan α cos β – sin β)

t(1 - \frac{sin α cos γ}{cos α sin γ}) = c (\frac{sin α cos β - cos α sin β}{cos α})\\ t(\frac{cos α sin γ-sin α cos γ}{cos α sin γ}) = c (\frac{sin (α-β)}{cos α})\\ t(\frac{sin(γ-α)}{cos α sin γ}) = c (\frac{sin (α-β)}{cos α})\\ t = c (\frac{sin (α-β)sin γ}{sin(γ-α)})

Now,

t+x = c (\frac{sin (α-β)sin γ}{sin(γ-α)}) + c sin β\\ = c (\frac{sin (α-β)sin γ}{sin(γ-α)} + sin β)\\ = c (\frac{sin (α-β)sin γ+sin(γ-α)sin β}{sin(γ-α)})\\ = c (\frac{sin γ(sin α cos β - cos α sin β)+sin β(sin γ cos α - cos γ sin α)}{sin(γ-α)})\\ = c (\frac{sin γ sin α cos β - sin γ cos α sin β + sin β sin γ cos α - sin β cos γ sin α}{sin(γ-α)})\\ = c (\frac{sin γ sin α cos β - sin β cos γ sin α}{sin(γ-α)})\\ = c (\frac{sin α (sin γ cos β - sin β cos γ)}{sin(γ-α)})\\ = c (\frac{sin α (sin(γ-β)}{sin(γ-α)})\\ = \frac{c sin α (sin(γ-β)}{sin(γ-α)}

Hence proved !!

Question 31: If the sides a, b and c of △ABC are in H.P. Prove thatsin^2\frac{A}{2}, sin^2\frac{B}{2}  andsin^2\frac{C}{2}  are in H.P

Solution:

If the sides a, b and c of △ABC are in H.P

Then,\frac{1}{a}  ,\frac{1}{b}  and\frac{1}{c}  are in AP

\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\\ \frac{a-b}{ba} = \frac{b-c}{ca}



According to the sine rule

\frac{a}{sin \hspace{0.1cm}A} = \frac{b}{sin \hspace{0.1cm}B} = \frac{c}{sin \hspace{0.1cm}C} = \lambda

\frac{sin \hspace{0.1cm}A-sin \hspace{0.1cm}B}{sin \hspace{0.1cm}Bsin \hspace{0.1cm}A} = \frac{sin \hspace{0.1cm}B-sin \hspace{0.1cm}C}{sin \hspace{0.1cm}Csin \hspace{0.1cm}B}

By using trigonometric formula,

sin a – sin b = 2 sin(\frac{a-b}{2})  cos(\frac{a+b}{2})

\frac{2 sin (\frac{A-B}{2}) cos(\frac{A+B}{2})}{sin \hspace{0.1cm}A} = \frac{sin (\frac{B-C}{2}) cos(\frac{B+C}{2})}{sin \hspace{0.1cm}C}\\ \frac{sin (\frac{A-B}{2}) cos(\frac{\pi-C}{2})}{sin \hspace{0.1cm}A} = \frac{sin (\frac{B-C}{2}) cos(\frac{\pi-A}{2})}{sin \hspace{0.1cm}C}\\ \frac{sin (\frac{A-B}{2}) cos(\frac{\pi}{2}-\frac{C}{2})}{sin \hspace{0.1cm}A} = \frac{sin (\frac{B-C}{2}) cos(\frac{\pi}-\frac{A}{2})}{sin \hspace{0.1cm}C}\\ \frac{sin (\frac{A-B}{2}) sin(\frac{C}{2})}{2 sin(\frac{A}{2})cos(\frac{A}{2})} = \frac{cos(\frac{C}{2}) sin(\frac{A}{2})}{2 sin(\frac{C}{2})cos(\frac{C}{2})}\\ (sin (\frac{A-B}{2}) sin(\frac{C}{2}))(2 sin(\frac{C}{2})cos(\frac{C}{2})) = 2 sin(\frac{A}{2})sin(\frac{A}{2}) (cos(\frac{A}{2}) cos(\frac{A}{2}))\\ sin (\frac{A-B}{2}) sin^2(\frac{C}{2}) cos(\frac{C}{2}) = cos(\frac{C}{2}) sin^2(\frac{A}{2})cos(\frac{A}{2})\\ sin (\frac{A-B}{2}) sin^2(\frac{C}{2}) sin(\frac{A+B}{2}) = cos(\frac{C}{2}) sin^2(\frac{A}{2})sin(\frac{B+C}{2})\\ sin^2(\frac{C}{2})(sin^2(\frac{A}{2})-sin^2(\frac{B}{2}))= sin^2(\frac{A}{2})(sin^2(\frac{B}{2})-sin^2(\frac{C}{2}))\\ sin^2(\frac{C}{2}) sin^2(\frac{A}{2}) - sin^2(\frac{C}{2}) sin^2(\frac{B}{2}))= sin^2(\frac{A}{2}) sin^2(\frac{B}{2})-sin^2(\frac{A}{2}) sin^2(\frac{C}{2})

Divide bysin^2\frac{A}{2} .sin^2\frac{B}{2}. sin^2\frac{C}{2}  , we get

\frac{1}{sin^2\frac{B}{2}} - \frac{1}{sin^2\frac{A}{2}} = \frac{1}{sin^2\frac{C}{2}} - \frac{1}{sin^2\frac{B}{2}}

Hence,sin^2\frac{A}{2}, sin^2\frac{B}{2}  andsin^2\frac{C}{2}  are in H.P

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




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