If in a triangle ABC #cosAcosB+sinAsinBsinC=1# then how will you prove that the triangle is right angled and isosceles?

2 Answers
Jun 27, 2016

Answer:

Please see below.

Explanation:

Multiplying both sides by #2# in given equality #cosAcosB+sinAsinBsinC=1#, we get

#2cosAcosB+2sinAsinBsinC=2# or

#2cosAcosB+2sinAsinBsinC=(sin^2A+cos^2A)+(sin^2B+cos^2B)#

or #(cos^2A+cos^2B-2cosAcosB)+(sin^2A+sin^2B-2sinAsinB)+2sinAsinB-2sinAsinBsinC=0# or

or #(cosA-cosB)^2+(sinA-sinB)^2+2sinAsinB(1-sinC)=0#

Note that all three terms contained above are positive, as while first two terms are squares and hence positive, third term is positive as sine of angles #A# and #B# is positive (as they are less than #pi# being angles of a triangle) and #(1-sinC)# too will be positive as #sinC<1#.

But, their sum is zero and hence each term is equal to zero, i.e.

#cosA-cosB=0#, #sinA-sinB=0# and #1-sinC=0# or

#cosA=cosB#, #sinA=sinB# and #1-sinC=0#

i.e. #A=B# and #sinC=1# i.e. #C=pi/2#

Hence the triangle is isosceles and right angled.

Jun 27, 2016

Answer:

as follows

Explanation:

Given
#cosAcosB+sinAsinBsinC=1#
#=>cosAcosB+sinAsinB-sinAsinB+sinAsinBsinC=1#

#=>cos(A-B)-sinAsinB(1-sinC)=1#

#=>1-cos(A-B)+sinAsinB(1-sinC)=0#
#=>2sin^2((A-B)/2)+sinAsinB(1-sinC)=0#

Now in above relation the first term being squared quantity will be positive.In the second term A,B and C all are less than
#180^@# but greater than zero.
So sinA ,sinB and sinC all are positive and less than 1.So the 2nd term as a whole is positive.
But RHS=0.
It is only possible iff each term becomes zero.

When #2sin^2((A-B)/2)=0#
then#A=B#

and when 2nd term=0 then
#sinAsinB(1-sinC)=0#

0< A and B <180
#=>sinA !=0and sinB!=0#

So #1-sinC=0=>C=pi/2#

So in triangle ABC
#A=B and C=pi/2->"the triangle is right angled and isosceles"#