Question

In a triangle abc if a=6 b=3 and cos(a-b)=4/5 the area of triangle is

Answer 1

Given:

a = 6

b = 3

Cos (a-b) = 4/5

To Find:

The area of triangle = ?

Solution:

As we know,

⇒  $tan(\frac{a-b}{2})$ = $\sqrt{\frac{1-cos(a-b)}{1+cos(a-b)}}$

On putting the estimated value, we get

                   = $\sqrt{\frac{1-\frac{4}{5} }{1+\frac{4}{5} } }$

                   = $\frac{1}{3}$

Also,

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

On substituting the given values, we get

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

⇒  $cot(\frac{c}{2})=1$

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

            $\frac{c}{2} =\frac{\pi}{4}$ i.e., 90°

Now,

The area of triangle will be:

= $\frac{1}{2}\times a\times b$

= $\frac{1}{2}\times 6\times 3$

= $9 \ sq \ units$

Thus, the area will be "9 sq units".

Answer 2

HELLO DEAR,

In a triangle ∆abc,

GIVEN:-

In a triangle ∆ABC

$\sf{a = 6, b = 3}$

$\sf{\cos ( a - b ) = \frac{4}{5}}$

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

$\sf{\rightarrow \tan\left(\frac{a-b}{2}\right) = }\sf{\sqrt{\frac{\frac{1}{\frac{5}{9}}}{5}}=} \sf{\frac{1}{3}}$

Therefore,

$\sf{\rightarrow \tan \left(\frac{a-b}{2}\right)=} \sf{\frac{1}{3}}$

Also we know:- $\sf{\tan\left(\frac{a-b}{2}\right) = } \sf{\frac{a-b}{a+b} \times \cot \left(\frac{c}{2}\right)}$

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

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

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

On comparing both side we get,

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

$\sf{ {\therefore, } c = \frac{\pi}{2}=90{\degree}}$

Therefore, the given triangle is a right angled triangle

$\sf{ar\triangle ABC = \frac{1}{2} \times a \times b}$

$\sf{Area\:of \:Triangle \:ABC \:=} \sf{ \frac{1}{2} \times 6 \times 3 = 9 { sq.\: units }}$

Hence, the area of triangle is 9 square units

I HOPE IT'S HELP YOU DEAR,

THANKS