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

If for a triangle ABC,
∣ ∣abcbcacab∣ ∣=0
then sin3A+sin3B+sin3C=

A
sinA+sinB+sinC
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3sinAsinBsinC
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
sin3A+sin3B+sin3C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
sin3Asin3Bsin3C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 3sinAsinBsinC
Δ=∣ ∣abcbcacab∣ ∣=(a+b+c)∣ ∣1bc1ca1ab∣ ∣
[applying C1C1+C2+C3 and taking (a+b+c) common from C1]=(a+b+c)∣ ∣1bc0cbac0abbc∣ ∣[applying R2R2R1 and R3R3R1]=(a+b+c)[(cb)(bc)(ab)(ac)]=(a+b+c)[ab+bc+caa2b2c2]=(a3+b3+c3)+3abc
Now, Δ=0 implies a3+b3+c3=3abc. Therefore, by the law of sines.
sinAa=sinBb=sinCc, we get
sin3A+sin3B+sin3C=3sinAsinBsinC.

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