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

Given:
sin(B+C)=32
sin(A+C)=sin B and
sin A=12

Find sin(A+BC)

A
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
32
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 32
Given,
sin(B+C)=32sin(A+C)=sin Bsin A=12

From sin A=12, we get A=30(1)
sin(B+C)=32 we know that B+C=60(2)
sin(A+C)=sinB we get A+C=B(3)
From the equations (1),(2) and (3)
C=60B, A=30
A+C=60B+30=B
2×B=90
B=45
B+C=60,C=6045
C=15
A+BC=30+4515=60
sin(A+BC)=sin 60=32

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