if ABC is a right angled triangle at B , sin(A) = 3/5 , then sin(B+c)

Question

if ABC is a right angled triangle at B , sin(A) = 3/5 , then sin(B+c)

Answer 1

To find sin(B + C), we need to first understand the values of angles B and C in a right-angled triangle.

In a right-angled triangle ABC, where angle B is the right angle, angle A is the angle opposite to side BC and angle C is the angle opposite to side AB.

The trigonometric relationship for right-angled triangles is given by:
sin(A) = opposite/hypotenuse.

In this case, we are given that sin(A) = 3/5. This means that the opposite side (BC) is 3 and the hypotenuse (AC) is 5.

We can use the Pythagorean theorem to find the length of the third side AB:
AB^2 = AC^2 - BC^2
AB^2 = 5^2 - 3^2
AB^2 = 25 - 9
AB^2 = 16
AB = 4

Now that we know the lengths of all three sides of the triangle, we can use the sine function to find the values of sin(B) and sin(C).

sin(B) = opposite/hypotenuse
sin(B) = AB/AC
sin(B) = 4/5

sin(C) = opposite/hypotenuse
sin(C) = BC/AC
sin(C) = 3/5

Now, let's find sin(B + C).

Using the sum of angles formula for sine, we have:
sin(B + C) = sin(B)cos(C) + cos(B)sin(C)

sin(B) = 4/5
cos(C) is adjacent/hypotenuse = BC/AC = 3/5
cos(B) is adjacent/hypotenuse = AB/AC = 4/5
sin(C) = 3/5

sin(B + C) = (4/5)(3/5) + (3/5)(4/5)
sin(B + C) = (12/25) + (12/25)
sin(B + C) = 24/25