Answer:

We can write the following set of equations:

θ₁ = θ₀ sin (wt + δ₁)                      ………………………….(1)

θ₂ = θ₀ sin (wt + δ₂)                     ……………………………..(2)

for first equation we have , θ = 2^o

Upon putting this values, the equation (1) reduces to –

wt + δ₁ = 90^o

and for the second equation, we have θ = -1^o

Similarly from (2), => wt + δ₂ = – 30^o

Therefore, upon equating te results we gte

δ₁ – δ₂ = 120^o