Solving this triangle using appropriate Law of Sine?

Hello friends.

I'm not well versed on how to make triangles on here. Please see the image.

enter image source here

Thanks all

1 Answer
Aug 1, 2018

#angle A=(36.87)^circ ,angle B=(36.87)^circ ,angle C=(106.26)^circ #

#" Area of triangle"=Delta=12 " sq. units"#

Explanation:

From the image :

#BC=CA=5 and AB=8#

#angleA=angleB=>color(green)(triangle ABC " is isosceles triangle."#

Also, #A+B+C=180^circ=>color(violet)(B and C " are acute angles."#

#=>a=5,b=5, andc=8#

#:."perimeter " : 2s=a+b+c=5+5+8=18#

#:.s=9=>s-a=4,s-b=4 and s-c=1#

#:." area of triangle"=Delta=sqrt(s(s-a)(s-b)(s-c))#

#=>color(brown)(Delta=sqrt(9(4)(4)(1))=sqrt144=12#

Now , we know that , # Delta=(abc)/(4R)=>R=(abc)/(4Delta)#

#:.R=(5xx5xx8)/(4xx12)=25/6#

#"Using "color(blue)"sine rule:"#

#color(blue)(a/sinA=b/sinB=c/sinC=2R#

#:.sinA=a/(2R)=5/(2xx(25/6))=30/50=0.6#

#:.angle A=arcsin(0.6)#

#=>color(red)(angleA=(36.87)^circ#

#=>color(red)(angleB=(36.87)^circ......to[becauseangleA=angleB]#

We have , #A+B+C=180^circ=>C=180^circ-(A+B)#

#:.C=180^circ-(36.87+36.87)^circ=180^circ-(73.74)^circ#

#:.color(red)(angle C=(106.26)^circ#

Hence,

#angle A=(36.87)^circ ,angle B=(36.87)^circ ,angle C=(106.26)^circ #

#" area of triangle"=Delta=12 " sq. units"#