Question

If A & B are acute angle such that sin A=4/5 and cos B=40/41 find sin (A+B)

Answer 1

Step-by-step explanation:

SIN (A+B)

sinA cosB + cosA sinB

4/5 × 40/41 + 9/41 × 3/5

Solve further and get Your answer

Answer 2

Answer:

Use the formula

☆ sin (A+B) = sinAcosB + sinBcosA ☆

Given :

• sin A = 4/5
• cos B = 40/41

From this, we can calculate :

cos A

Opposite side = 4

Hypotenuse = 5

Therefore, adjacent side =

$\sqrt{25 - 16} = \sqrt{9 \: } = 3$

By Pythagoras theorem

Therefore, cosA = 3/5

sinB

Adjacent side = 40

Hypotenuse = 41

Opposite side :

$\sqrt{1681 - 1600} = \sqrt{81} = 9$

sinB = 9/41

Substituting in formula, we get

sin (A+B) =

$\frac{4}{5} \times \frac{40}{41} + \frac{3}{ 5} \times \ \frac{9}{41}$

=

$\frac{187}{205}$