Question

Please help. My so called "friend" didnt want to help.

Answer 1

$-\sqrt{2}sec(x) = 2$ {0° ≤ x ≤ 360°}
$\frac{-\sqrt{2}sec(x)}{-\sqrt{2}} = \frac{2}{-\sqrt{2}}$
$sec(x) = -\frac{2}{\sqrt{2}}$
$sec(x) = -\frac{2}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$
$sec(x) = -\frac{2\sqrt{2}}{2}$
$sec(x) = -\sqrt{2}$
$sec^{-1}(sec(x)) = sec^{-1}(-\sqrt{2})$
$x = \frac{1}{cos(-\sqrt{2})}$
$x = \frac{1}{cos^{-1}(\sqrt{2})}$
$x = sec^{-1}(\sqrt{2})$
$x = 135\°, 225\°$
The answer is 3.

   c² = a² + b²
 17² = 8² + b²
289 = 64 + b²
- 64  - 64
225 = b²
  15 = b
$cos(x) = \frac{15}{17} =$
$cos^{-1}(cos(x)) = cos^{-1}(\frac{15}{17})$
$x = 28.07248694$
$28.07248694 = 28\°4'20.95" \approx 28\°4'$
The answer is 2.