If in a triangle ABC, AB = 5 units, [ B = ^ - 1( 3 5 )] and radius of circumcircle of []AB…

Question

If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is :

Accepted Answer

As, $$\cos B = {3 \over 5} \Rightarrow B = 53^\circ $$

As, $$R = 5 \Rightarrow {c \over {\sin c}} = 2R$$

$$ \Rightarrow {5 \over {10}} = \sin c \Rightarrow C = 30^\circ $$

Now, $${b \over {\sin B}} = 2R \Rightarrow b = 2(5)\left( {{4 \over 5}} \right) = 8$$

Now, by cosine formula

$$\cos B = {{{a^2} + {c^2} - {b^2}} \over {2ac}}$$

$$ \Rightarrow {3 \over 5} = {{{a^2} + 25 - 64} \over {2(5)a}}$$

$$ \Rightarrow {a^2} - 6a - 3g = 0$$

$$\therefore$$ $$a = {{6 \pm \sqrt {192} } \over 2} = {{6 \pm 8\sqrt 3 } \over 2}$$

$$ \Rightarrow 3 + 4\sqrt 3 $$ (Reject $$a = 3 - 4\sqrt 3 $$)

Now, $$\Delta  = {{abc} \over {4R}} = {{(3 + 4\sqrt 3 )(8)(5)} \over {4(5)}} = 2(3 + 4\sqrt 3 )$$

$$ \Rightarrow \Delta  = (6 + 8\sqrt 3 )$$

$$\Rightarrow$$ Option (3) is correct.

Solutions · 1