Question

12. (05.02 LC) Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = 7 over q and tan y° = 7 over r, what is the value of sec y°? (4 points) sec y° = q over r sec y° = 7r sec y° = 7q sec y° = r over q

Answer 1

Answer:

$sec~y=\frac{q}{r}$

Step-by-step explanation:

$tan ~y=\frac{7}{r} \\\\\frac{sin~y}{cos~y} =\frac{7}{r} \\\\sin~y=\frac{7}{r} cos~y\\sin ~y=\frac{7}{q} \\\frac{7}{q} =\frac{7}{r} cos~y\\sec~y=\frac{7}{r} \times \frac{q}{7} =\frac{q}{r}$