Answer:
√29/5
Explanation:
The terminal ray of ∠A passes through the point (10,−4).
From the diagram above, angle A is in quadrant IV.
First, find the value of r using the Pythagoras theorem:
[tex]\begin{gathered} r^2=10^2+(-4)^2 \\ r^2=100+16 \\ r^2=116 \\ r=\sqrt{116} \\ r=2\sqrt{29} \end{gathered}[/tex]Secant is the inverse of cosine.
Therefore:
[tex]\begin{gathered} \sec A=\frac{Hypotenuse}{Adjacent} \\ =\frac{r}{x} \\ =\frac{2\sqrt{29}}{10} \\ \sec A=\frac{\sqrt{29}}{5} \end{gathered}[/tex]The value of sec A is √29/5.
Method 2
The terminal ray of ∠A passes through the point (10,−4).
[tex]\begin{gathered} (x,y)=(10,-4) \\ \text{ Opposite Side, }y=-4 \\ \text{ Adjacent Side, }x=10 \end{gathered}[/tex]Find the value of the hypotenuse, r using the Pythagoras theorem:
[tex]\begin{gathered} r^2=10^2+(-4)^2 \\ r^2=100+16 \\ r^2=116 \\ r=\sqrt{116} \\ r=2\sqrt{29} \end{gathered}[/tex]Secant is the inverse of cosine.
Therefore:
[tex]\begin{gathered} \sec A=\frac{Hypotenuse}{Adjacent} \\ =\frac{r}{x} \\ =\frac{2\sqrt{29}}{10} \\ \sec A=\frac{\sqrt{29}}{5} \end{gathered}[/tex]The value of sec A is √29/5.
