Use the ff to find each trigonometric Ratio /// BRAINLYS TO WHOEVER ANSWERS THIS /// Please no nonsense Answers.
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Use the ff to find each trigonometric Ratio /// BRAINLYS TO WHOEVER ANSWERS THIS /// Please no nonsense Answers.
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Given the triangle PQR, given side PR = 13cm, and side QR = 12, we find the length of the side r or side PQ.
Based on the diagram, triangle PQR is a right triangle. With that, we have hypotenuse (side PR) and the legs (side PQ and QR). We can then use pythagorean formula to solve for the leg PQ:
[tex] PQ^2 + QR^2 = PR^2 [/tex]
[tex] PQ^2 + 12\textsf{cm}^2 = 13\textsf{cm}^2 [/tex]
[tex] PQ^2 + 144 = 169 [/tex]
[tex] PQ^2 = 169 - 144 [/tex]
[tex] PQ^2 = 25 [/tex]
[tex] \sqrt{PQ^2} = \sqrt{25} [/tex]
[tex] \boxed{PQ = 5\textsf{ cm}} [/tex]
Now we have PQ = 5cm, QR = 12cm, PR = 13cm, lets solve and find the trigonometric ratios for the triangle PQR:
The formula for trigonometric ratios are:
[tex] \displaystyle \textsf{sin }\theta= \frac{\textsf{opposite}}{\textsf{hypotenuse}} [/tex]
[tex] \displaystyle \textsf{cos }\theta = \frac{\textsf{adjacent}}{\textsf{hypotenuse}} [/tex]
[tex] \displaystyle \textsf{tan }\theta = \frac{\textsf{opposite}}{\textsf{adjacent}} [/tex]
[tex] \displaystyle \textsf{csc }\theta = \frac{\textsf{1}}{\textsf{sin}} [/tex]
[tex] \displaystyle \textsf{sec }\theta = \frac{\textsf{1}}{\textsf{cos}} [/tex]
[tex] \displaystyle \textsf{cot }\theta = \frac{\textsf{1}}{\textsf{tan}} [/tex]
Answer:
[tex] \displaystyle \sin R = \frac{5}{13} [/tex]
[tex] \displaystyle \cos R = \frac{12}{13} [/tex]
[tex] \displaystyle \tan R = \frac{5}{12} [/tex]
[tex] \displaystyle \sin P = \frac{12}{13} [/tex]
[tex] \displaystyle \cos P = \frac{5}{13} [/tex]
[tex] \displaystyle \tan P = \frac{12}{5} [/tex]
[tex] \displaystyle \csc R = \frac{13}{5} [/tex]
[tex] \displaystyle \sec R = \frac{13}{12} [/tex]
[tex] \displaystyle \cot R = \frac{12}{5} [/tex]
[tex] \displaystyle \csc P = \frac{13}{12} [/tex]
[tex] \displaystyle \sec P = \frac{13}{5} [/tex]
[tex] \displaystyle \cot P = \frac{5}{12} [/tex]