Answer
$1$
Work Step by Step
We know that $\sin.(\theta)=\cos(90^{o}-\theta)$.
Hence,
$\dfrac{\cos(10^{o})}{\sin(80^{o})}
\\=\dfrac{\cos(10^{o})}{\cos(90^{o}-80^{o})}
\\=\dfrac{\cos(10^{o})}{\cos(10^{o})}
\\=1$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 8 - Applications of Trigonometric Functions - 8.1 Right Triangle Trigonometry; Applications - 8.1 Assess Your Understanding - Page 519: 21
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