Question

In a triangle abc, a=6,b=3 and cos(a-b)=4/5, find area of triangle.

Answer 1

In a triangle abc, a=6,b=3 and cos(a-b)=4/5 then the area of triangle is 9 square units

Solution:

Given:- Triangle ABC whose values are:-

$a=6, b=3 \text { and } \cos (a-b)=\frac{4}{5}$

We know that,

$\tan \left(\frac{a-b}{2}\right)=\sqrt{\frac{1-\cos (a-b)}{1+\cos (a-b)}}=\sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}$

$=\sqrt{\frac{\frac{1}{\frac{5}{9}}}{5}}=\frac{1}{3}$

$\text {Therefore, } \tan \left(\frac{a-b}{2}\right)=\frac{1}{3}$

Also we know that:

$\tan \left(\frac{a-b}{2}\right)=\frac{a-b}{a+b} \times \cot \left(\frac{c}{2}\right)$

$\frac{1}{3}=\frac{6-3}{6+3} \times \cot \left(\frac{c}{2}\right)$

$\frac{1}{3}=\frac{3}{9} \times \cot \left(\frac{c}{2}\right)$

$\cot \left(\frac{c}{2}\right)=1=\cot \left(\frac{\pi}{4}\right)$

On comparing we get,

$\frac{c}{2}=\frac{\pi}{4}$

$\text {Therefore, } c=\frac{\pi}{2}=90 \text { degree }$

Therefore, the given triangle is a right angled triangle

$\text {Area of Triangle }=\frac{1}{2} \times a \times b$

$\text {Area of Triangle }=\frac{1}{2} \times 6 \times 3=9 \text { sq units }$

Thus the area of triangle is 9 square units

Learn more about area of triangle

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Answer 2

Answer:

Step-by-step explanation:

R.E.F image  

Given that a=6,b=3 & cos(A−B)=4/5

Solution :- We know that  

tan(  

2

A−B

​  

)=  

1+cos(A−B)

1−cos(A−B)

​  

 

​  

 

=  

1+4/5

1−4/5

​  

 

​  

 

=  

5+4

5−4

​  

 

​  

=  

9

1

​  

 

​  

=  

3

1

​  

 

∴tan(  

2

A−B

​  

)=  

3

1

​  

 

mow, we know that,  

tan(  

2

A−B

​  

)=  

a+b

a−b

​  

cot  

2

c

​  

 

3

1

​  

=  

6+3

6−3

​  

cot  

2

c

​  

 

cot  

2

c

​  

=1=cot  

4

π

​  

 

c=  

4

2π

​  

=  

2

π

​  

 (right angle)  

now, it is right angle triangle  

so,  

Area of ΔABC=  

2

1

​  

×base×height

=  

2

1

​  

×a×b

=  

2

1

​  

×6×3

=9sq.units