Triangle BON, DOM, ADB, ACB and CDB are similar triangles.

∠ACB = θ


For triangle BON ,

sin θ = opposite/hypotenuse = OB/BN = BN / 6

BN = 6 sin θ

For triangle ACB,

cos θ = adjacent / hypotenuse = AC / AB = AC / 6

AC = 2 cos θ

AM
= CM + AC
= BN + AC (since CBMN is a rectangle, CM = BN)
= 6 sin θ + 2 cos θ

But now, look at triangle AOM.

sin ∠AOM = opposite/hypotenuse = AM / OA

sin (θ + α) = (6 sin θ + 2 cos θ) / OA

OA sin (θ + α) = 6 sin θ + 2 cos θ


How to find OA? Look at triangle AOB.

OA² = OB² + AB² (Pythagoras's Theorem)

OA² = 6² + 2²

OA = √(6² + 2²) = √40

How to find α ?

tan α = opposite / adjacent = AB/OB = 2/6 = ⅓

α = tan-¹ (⅓)
(The basic angle)


So, putting these together,

√(6² + 2²) sin (θ + tan-¹(⅓) ) = 6 sin θ + 2 cos θ

√40 sin (θ + tan-¹(⅓)) = 6 sin θ + 2 cos θ

Sorry, ∠CAB = θ, not ∠ACB

Thx for clearing my doubts!! I got it now

No prob