English

In Triangle Abc, Prove the Following: a + B C = Cos ( a − B 2 ) Sin C 2 - Mathematics

Advertisements
Advertisements

Question

In triangle ABC, prove the following: 

\[\frac{a + b}{c} = \frac{\cos \left( \frac{A - B}{2} \right)}{\sin \frac{C}{2}}\]

 

Advertisements

Solution

Let 

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]    ...(1) 

Then,
Consider the LHS of the equation

\[\frac{a + b}{c} = \frac{\cos \left( \frac{A - B}{2} \right)}{\sin \frac{C}{2}}\]

\[LHS = \frac{a + b}{c}\]
\[ = \frac{k\sin A + k\sin B}{k\sin C} \left( using \left( 1 \right) \right)\]
\[ = \frac{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{2\sin\frac{C}{2}\cos\frac{C}{2}}\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\cos\left( \frac{\pi - \left( A + B \right)}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\sin\left( \frac{A + B}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}} = RHS \]
\[\text{ Hence proved } .\]

 

shaalaa.com
Sine and Cosine Formulae and Their Applications
  Is there an error in this question or solution?
Chapter 10: Sine and cosine formulae and their applications - Exercise 10.1 [Page 13]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 8 | Page 13

RELATED QUESTIONS

If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides. 


In triangle ABC, prove the following:

\[\frac{c}{a - b} = \frac{\tan\left( \frac{A}{2} \right) + \tan \left( \frac{B}{2} \right)}{\tan \left( \frac{A}{2} \right) - \tan \left( \frac{B}{2} \right)}\]

 


In any triangle ABC, prove the following: 

\[\sin \left( \frac{B - C}{2} \right) = \frac{b - c}{a} \cos\frac{A}{2}\]

 


In triangle ABC, prove the following: 

\[a^2 \sin \left( B - C \right) = \left( b^2 - c^2 \right) \sin A\]

 


In triangle ABC, prove the following: 

\[\frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{\sin B}} = \frac{a + b - 2\sqrt{ab}}{a - b}\]

 


In triangle ABC, prove the following: 

\[a \left( \sin B - \sin C \right) + \left( \sin C - \sin A \right) + c \left( \sin A - \sin B \right) = 0\]

 


In triangle ABC, prove the following: 

\[a^2 \left( \cos^2 B - \cos^2 C \right) + b^2 \left( \cos^2 C - \cos^2 A \right) + c^2 \left( \cos^2 A - \cos^2 B \right) = 0\]

 


\[a \left( \cos B \cos C + \cos A \right) = b \left( \cos C \cos A + \cos B \right) = c \left( \cos A \cos B + \cos C \right)\]


In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ). 


In ∆ABC, if sin2 A + sin2 B = sin2 C. show that the triangle is right-angled. 


In ∆ABC, if a2b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P. 


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


If the sides ab and c of ∆ABC are in H.P., prove that \[\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2} \text{ and } \sin^2 \frac{C}{2}\]


The sides of a triangle are a = 4, b = 6 and c = 8. Show that \[8 \cos A + 16 \cos B + 4 \cos C = 17\]


In ∆ ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C


In ∆ABC, prove the following: \[b \left( c \cos A - a \cos C \right) = c^2 - a^2\]


In ∆ABC, prove the following: \[c \left( a \cos B - b \cos A \right) = a^2 - b^2\]


In ∆ABC, prove  the following: 

\[2 \left( bc \cos A + ca \cos B + ab \cos C \right) = a^2 + b^2 + c^2\]

 


In ∆ABC, prove the following:

\[\frac{c - b \cos A}{b - c \cos A} = \frac{\cos B}{\cos C}\] 

 


In \[∆ ABC, \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}\]  Prove that \[\frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}\] 


In \[∆ ABC, if \angle B = 60°,\]  prove that \[\left( a + b + c \right) \left( a - b + c \right) = 3ca\]


Answer  the following questions in one word or one sentence or as per exact requirement of the question. 

Find the area of the triangle ∆ABC in which a = 1, b = 2 and \[\angle C = 60º\] 



Answer  the following questions in one word or one sentence or as per exact requirement of the question. 

In a ∆ABC, if b = 20, c = 21 and \[\sin A = \frac{3}{5}\] 

 


Answer the following questions in one word or one sentence or as per exact requirement of the question.  

In ∆ABC, if a = 8, b = 10, c = 12 and C = λA, find the value of λ


Answer the following questions in one word or one sentence or as per exact requirement of the question. 

In any ∆ABC, find the value of

\[\sum^{}_{}a\left( \text{ sin }B - \text{ sin }C \right)\]


Mark the correct alternative in each of the following:
In any ∆ABC, \[\sum^{}_{} a^2 \left( \sin B - \sin C \right)\] = 


Mark the correct alternative in each of the following: 

In a ∆ABC, if a = 2, \[\angle B = 60°\]  and\[\angle C = 75°\] 

 


Mark the correct alternative in each of the following:
If the sides of a triangle are in the ratio \[1: \sqrt{3}: 2\] then the measure of its greatest angle is 


Mark the correct alternative in each of the following: 

In a triangle ABC, a = 4, b = 3, \[\angle A = 60°\]   then c is a root of the equation 


Mark the correct alternative in each of the following: 

In a ∆ABC, if  \[\left( c + a + b \right)\left( a + b - c \right) = ab\] then the measure of angle C is 


Mark the correct alternative in each of the following:

In any ∆ABC, \[a\left( b\cos C - c\cos B \right) =\]  


Find the value of `(1 + cos  pi/8)(1 + cos  (3pi)/8)(1 + cos  (5pi)/8)(1 + cos  (7pi)/8)`


If x = sec Φ – tan Φ and y = cosec Φ + cot Φ then show that xy + x – y + 1 = 0
[Hint: Find xy + 1 and then show that x – y = –(xy + 1)]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×