### Video Transcript

Find sin of 𝐵 given tan of 𝐵 is
equal to four-thirds and cos of 𝐵 is equal to three-fifths.

In this question, we are told that
both the tangent and cosine of our angle are positive. From our CAST diagram, this means
that the angle lies in the first quadrant, as this is the only quadrant where the
tangent and cosine of an angle are both positive. This means that 𝐵 is greater than
zero degrees and less than 90 degrees. We know that the sine of any angle
in the first quadrant must also be positive. sin of angle 𝐵 must be greater
than zero.

As cos of 𝐵 equals three-fifths
and tan of 𝐵 equals four-thirds, we can sketch a right triangle in the first
quadrant as shown. This triangle is a Pythagorean
triple consisting of three positive integers, three, four, and five, such that three
squared plus four squared is equal to five squared. We know that in a right triangle,
the sin of any angle 𝜃 is equal to the opposite over the hypotenuse. The sin of angle 𝐵 is therefore
equal to four-fifths. If tan of 𝐵 is equal to
four-thirds and cos of 𝐵 is equal to three-fifths, then sin of 𝐵 is equal to
four-fifths.

An alternative method here would be
to recall one of our trigonometric identities. tan of 𝜃 is equal to sin of 𝜃
over cos of 𝜃. Multiplying both sides by cos of
𝜃, we have cos of 𝜃 times tan of 𝜃 is equal to sin of 𝜃. Replacing the angle 𝜃 with our
angle 𝐵, we can now substitute in the values for tan of 𝐵 and cos of 𝐵. This gives us sin of 𝐵 is equal to
three-fifths multiplied by four-thirds. Dividing the numerator and
denominator by a common factor of three, we are left with one-fifth multiplied by
four over one. This confirms our answer that sin
of 𝐵 is equal to four-fifths.