wiz-icon
MyQuestionIcon
MyQuestionIcon
9
You visited us 9 times! Enjoying our articles? Unlock Full Access!
Question

In Δ ABC the sides opposite to angles A,B,C are denoted by a,b,c respectively, then acos(BC)+bcos(CA)+ccos(AB) is equal to?

A
abcR2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
abc4R2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4abcR2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
abc2R2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A abcR2
Formula used:
Sine Rule; asinA=bsinB=csinC=2R
Projection rule; c=acosB+bcosA
We have;
acos(BC)+bcos(CA)+ccos(AB)
=a(cosB.cosC+sinB.sinC)+b(cosC.cosA+sinCsinA)+c(cosA.cosB+sinA.sinB)
Putting sinA=a2R;sinB=b2R and sinC=c2R
=acosB.cosC+bcosC.cosA+ccosA.cosB+3abc4R2
=cosC(acosB+bcosA)+ccosA.cosB+3abc4R2
By projection rule (acosB+bcosA)=c
=cosC.c+ccosA.cosB+3abc4R2
=c(cos[π(A+B)]+cosA.cosB)+3abc4R2
=c(cos(A+B)+cosA.cosB)+3abc4R2
=c.sinA.sinB+3abc4R2
Using Sine rule;
sinA=a2R,sinB=b2R

=abc4R2+3abc4R2=abcR2
Hence;
acos(BC)+bcos(CA)+Ccos(AB)=abcR2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Principal Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon