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

Find the value of θ which satisfy rsinθ=3 and r=4(1+sinθ),0θ2π.

Open in App
Solution

Given that

rsinθ=3 ....(1)

r=4(1+sinθ) ....(2)

From (1) we have sinθ=3r

Substitute this in (2), we get

r=4+4×(3r)

r=4+12r

r2=4r+12

r24r12=0

(r6)(r+2)=0

r=6,r=2

r=6sinθ=36=12

So, θ=π6,5π6

r=2sinθ=32 which is not possible.

Therefore, θ=π6,5π6

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