Question 1137926
A vertices of a triangle A({{{1}}},{{{2}}}) B({{{4}}},{{{5}}}) C({{{3}}},{{{-2}}}).


(a) find the surd form the length AB, BC and AC


A({{{1}}},{{{2}}}) B({{{4}}},{{{5}}})

{{{AB=sqrt((4-1)^2+(5-2)^2)}}} 

{{{AB=sqrt(3^2+3^2)}}} 

{{{AB=sqrt(2*3^2)}}} 

{{{AB=sqrt(18)}}} 

{{{AB=3sqrt(2)}}} 


 B({{{4}}},{{{5}}}) C({{{3}}},{{{-2}}})

{{{BC=sqrt((3-4)^2+(-2-5)^2)}}} 

{{{BC=sqrt((-1)^2+(-7)^2)}}} 

{{{BC=sqrt(1+49)}}}

{{{BC=sqrt(50)}}}

{{{BC=5sqrt(2)}}}


A({{{1}}},{{{2}}})  C({{{3}}},{{{-2}}})

{{{AC=sqrt((3-1)^2+(-2-2)^2)}}} 

{{{AC=sqrt(2^2+(-4)^2)}}} 

{{{AC=sqrt(4+16)}}} 

{{{AC=sqrt(20)}}} 
 
{{{AC=2sqrt(5)}}} 


(b) show that{{{ cos(A)=-1/sqrt(10)}}} and deduce the value of {{{sin(A)}}} in surd form 

{{{cos(A) =  (b^2 + c^2- a^2)/(2bc)}}}

use
{{{AB=c=sqrt(18)}}} 

{{{BC=a=sqrt(50)}}}

{{{AC=b=sqrt(20)}}} ..will be easier


{{{cos(A) =  ((sqrt(20))^2 + (sqrt(18))^2-(sqrt(50))^2)/(2sqrt(20)*sqrt(18))}}}

{{{cos(A) =  (20 + 18-50)/(12sqrt(10))}}}

{{{cos(A) =  (-12)/(12sqrt(10))}}}

{{{cos(A) =  (-cross(12)1)/(cross(12)sqrt(10))}}}

{{{cos(A) =  -1/sqrt(10)}}}

{{{sin(A)}}}=..........since {{{sin^2(A)=1-cos^2(A) =1-(1/sqrt(10))^2=1-1/10=9/10}}}=>

{{{sin(A)=sqrt(9/10)}}}
{{{sin(A)=sqrt(9)/sqrt(10)}}}
{{{sin(A)=3/sqrt(10)}}}

{{{sin(A)=3sqrt(10)/10}}}