Question

question 14 (1 point)
what is the measure of da, if \\( \sec a = \sqrt { 2 } \\) and da is located in the first quadrant?
\\( \bigcirc \\) a) \\( 135 ^ { \circ } \\)
\\( \bigcirc \\) b) \\( 45 ^ { \circ } \\)
\\( \bigcirc \\) c) \\( 70.5 ^ { \circ } \\)
\\( \bigcirc \\) d) \\( 22.5 ^ { \circ } \\)

Explanation:

Step1: Recall the secant - cosine relationship

We know that \(\sec A=\frac{1}{\cos A}\). Given \(\sec A = \sqrt{2}\), then \(\cos A=\frac{1}{\sqrt{2}}\).

Step2: Find the angle \(A\) using the inverse - cosine function

If \(\cos A=\frac{1}{\sqrt{2}}\), then \(A = \cos^{-1}(\frac{1}{\sqrt{2}})\).
We know that \(\cos45^{\circ}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\) and the range of the inverse - cosine function \(y = \cos^{-1}(x)\) is \([0,\pi]\) (or \([0^{\circ},180^{\circ}]\)). Since the angle \(A\) is in the first quadrant (\(0^{\circ}

Answer:

B. \(45^{\circ}\)