IIT JEE , AIEEE Forum - Mathematics , Trignometry

ashish_banga

in a triangle ABC , A = 120 , b - c = 3 sq rt 3 and area of triangle = 4.5 sq rt3 . then a is

pink_ele

here
area=4.5 rt3
1/2bcsinA=4.5rt
bc=18
now already u have b-c=3rt3
itgives
b=3/2(rt3+rt11)
now u can get c as well:)

allamraju

pink_ele

yeah it is rt3/2..........thats why in d sol....rt 11is coming meas .........do it by shreedharacharyas rule dear

thedumbheadwithnobrain

$\frac{1}{2}bc(sinA)=\frac{9\sqrt{3}}{2}$

$\frac{1}{2}bc\frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{2}$

$(b-c)^2=27\Rightarrow b^2+c^2-2bc=27$

$cosA=\frac{-a^2+b^2+c^2}{2bc}$

$\frac{1}{2}=\frac{27+36-a^2}{36}$

sry i have edited this

allamraju