Question

Suppose ø is an angle in the standard position whose terminal side is in Quadrant I and tan ø = 60/11

Answer 1

Answer:

Following are the ratio to this question:

Step-by-step explanation:

Given:

$\bold{\tan \phi = \frac{60}{11} }$

Formula:

$\sin \phi = \frac{Perpendicular}{Hypotenuse}\\\\\cos \phi = \frac{Base}{Hypotenuse} \\\\\tan \phi = \frac{Perpendicular}{Base} \\\\\ cosec \phi = \frac{Hypotenuse}{Perpendicular}\\\\\sec \phi =\frac{Hypotenuse}{Base}\\\\\cot \phi = \frac{Base}{Perpendicular}$

$\ Base = 11 \\\ Perpendicular =60\\\ Hypotenuse= 61$

$\sin \phi = \frac{60}{61}\\\\\cos \phi = \frac{11}{61} \\\\\tan \phi = \frac{60}{11} \\\\\ cosec \phi = \frac{ 61}{60}\\\\\sec \phi =\frac{61}{11}\\\\\cot \phi = \frac{11}{60}$