Give each trig ratio as a fraction in its simplest form.
Sin Q = Cos Q = Tan Q = Sin R = Cos R= Tan R=​

Answer:

sin q = 14/50

cos q = 24/25

tan q = 7/24

sin r = 24/25

cos r = 7/25

tan r = 24/7

Step-by-step explanation:

The required trigonometric ratio are as follows;

[tex]SinR= \frac{24}{25}, CosR = \frac{7}{25}, TanR = \frac{24}{7} \\\\SinQ = \frac{7}{25}, CosQ = \frac{24}{25}, TanQ = \frac{7}{24}[/tex]

WE have to determine the trigonometric function in simplest form.

In order to find trigonometric ratio, In right angle triangle QRP.

Length of PR = 14

Length of QR = 50

To find length of PQ using Pythagoras theorem.

[tex](H)^{2} = (P)^{2} + (B)^{2}\\\\(50)^{2} = (14)^{2} + (B)^{2} \\\\B^{2} = 2500 - 196\\\\B^{2} = 2304\\\\B = 48[/tex]

Length of PQ is 48.

Therefore,

Sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse)

Then,

[tex]SinQ = \frac{perpendicular}{hypotenuse} \\\\SinQ = \frac{14}{50}\\\\SinQ = \frac{7}{25}[/tex]

The ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

[tex]CosQ = \frac{Base}{hyptotenuse}\\\\CosQ = \frac{48}{50} \\\\CosQ = \frac{24}{25}[/tex]

The tangent of an angle is the length of the opposite side divided by the length of the adjacent side.

[tex]TanQ = \frac{perpendicular}{Base}\\\\TanQ = \frac{14}{48} \\\\TanQ = \frac{7}{24}[/tex]

And In right angle triangle RPQ,

Sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse).

[tex]SinQ = \frac{perpendicular}{hypotenuse} \\\\SinQ = \frac{48}{50}\\\\SinQ = \frac{24}{25}[/tex]

The ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

[tex]CosQ = \frac{Base}{hyptotenuse}\\\\CosQ = \frac{14}{50} \\\\CosQ = \frac{7}{25}[/tex]

The tangent of an angle is the length of the opposite side divided by the length of the adjacent side.

[tex]TanQ = \frac{perpendicular}{Base}\\\\TanQ = \frac{48}{14} \\\\TanQ = \frac{24}{7}[/tex]

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