Question

3. Find sin A and cos B exactly.
B
a
C
6
4
A

Answer 1

Check the picture below.

`\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies √(c^2-b^2)=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ √(6^2 - 4^2)=a\implies √(36-16)=a\implies √(20)=a\implies 2√(5)=a \\\\[-0.35em] ~\dotfill\\\\ sin(A )=\cfrac{\stackrel{opposite}{2√(5)}}{\underset{hypotenuse}{6}}\implies sin(A)=\cfrac{√(5)}{3} \\\\\\ cos(B )=\cfrac{\stackrel{adjacent}{2√(5)}}{\underset{hypotenuse}{6}}\implies cos(B)=\cfrac{√(5)}{3}`

Answer 2

The calculated value of sin B and cosine B is √5/3

How to determine the sin and the cosine of the angles

From the question, we have the following parameters that can be used in our computation:

The right triangle

Where, we have

cos(A) = sin(B)

Also, we have

cos(A) = Adjacent/Hypotenuse

In this case, the adjacent using the Pythagoras theorem is

Adjacent = √[6² - 4²]

Adjacent = √20

Adjacent = 2√5

Substitute the known values into the equation

cos(A) = sin(B) = 2√5/6

Simplify

cos(A) = sin(B) = √5/3

Hence, the sin and the cosine of the angles is √5/3