Question

What is the trigonometric ratio for sin S ?

Express your answer, as a simplified fraction.

Answer 1

$QR=\sqrt{68^2-60^2}\\QR=\sqrt{4624-3600}\\QR=\sqrt{1024}\\QR=32\\\\sinS=\frac{32}{68}\\sinS=\frac{8}{17}$

Answer 2

For this case, we have that by definition:

Be a rectangular triangle and an "x" angle.

$Sine (x) = \frac {CO} {H}$

CO is the leg opposite the angle and H the hypotenuse

We want to find the Sine (S) according to the figure shown:

$Sine {S} = \frac {QR} {68}$

We do not have the opposite leg, we must apply the Pythagorean theorem, which states:

$H = \sqrt {(CO) ^ 2 + (CA) ^ 2}$

Where:

H: Hypotenuse

CO: Opposite leg

CA: Adjacent leg

In this case, we must find CO:

$CO = \sqrt {H ^ 2- (CA) ^ 2}$

Where:

$H = 68\\CA = 60$

Substituting:

$CO = \sqrt {68 ^ 2-60 ^ 2}\\CO = \sqrt {4624-3600}\\CO = \sqrt {1024}\\CO = 32$

So, we have:

$Sine (S) = \frac {32} {68}$

Answer:

$Sine (S) = \frac {32} {68}$

$Sine (S) = \frac {8} {17}$