wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a triangle ABC, if cos(AC)cosB+cos2B=0 then a2,b2,c2 are in

A
H.P
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
G.P
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
A.P
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C A.P
cos(AC)cosB=cos2B
cos(AC)cos(π(A+C))=cos2B
cos(AC)cos(A+C)=cos2B
cos(AC)cos(A+C)=cos2B
cos2Asin2C=12sin2B
1sin2Asin2C=12sin2B
sin2A+sin2C=2sin2B
a2+c2=2b2 by sine rule.
Hence a2,b2,c2 are in A.P.

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