Math, asked by viveksathyan7977, 1 month ago

sinA = 3/5 and cosB =9/41 and A and B lie in the first quadrant, find sin (A+B) .

Answers

Answered by sharanyalanka7
5

Answer:

187/205

Step-by-step explanation:

Given,

sinA = 3/4

cosB = 9/41

A , B lie in first quadrant.

To Find :-

Value of 'sin(A + B)'

How To Do :-

As they gave that A , B lie in the first quadrant i.e in first quadrant all ratios are positive . To find the value of 'cosA and sinB' we need to find the remaining side of them by using pythagoreas theorem and by using the sin(A+B) formula we will get the value of that.

Formula Required :-

Pythogoreas theorem :-

(Hypotenuse side)² = (opposite side)² + (adjacent side)²

sinα = opposite side/hypotenuse side

cosα = adjacent side/hypotenuse side

sin(A + B) = sinAcosB + cosAsinB

Solution :-

sinA = 3/5

→ opposite side/hypotenuse side = 3/5

opposite side = 3 , hypotenuse side = 5

Let,

Adjacent side be 'x'

Applying pythogoreas theorem :-

(5)² = 3² + x²

25 = 9 + x²

25 - 9 = x²

16 = x²

x = √16

x = 4

∴ Adjacent side = x = 4

cosA = adjacent side/hypotenuse side

= 4/5

∴ cosA = 4/5

cosB = 9/41

adjacent side/hypotenuse side = 9/41

adjacent side = 9 , hypotenuse side = 41

Let, opposite side be 'x'

Applying pythagoreas theorem : -

(41)² = 9² + x²

1681 = 81 + x²

1681 - 81 = x²

x² = 1600

x =√1600

x = 40

∴ Opposite side = x = 40

sinB = opposite side/hypotenuse side

∴sinB = 40/41

sin(A + B) = sinAcosB + sinBcosA

= 3/5 × 9/41 + 4/5 × 40/41

= 27/205 + 160/205

= (27 + 160)/205

= 187/205

∴ sin(A + B) = 187/205

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