Question

The Answer to This Question

Answer 1

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Answer 2

Recall the Trigonometric Ratio below:

$\tt{ \large{sin \theta = \frac{opposite}{hypotenuse} } }\\ \tt{\large{cos \theta = \frac{adjacent}{hypotenuse} }} \\ \tt{ \large{tan \theta = \frac{opposite}{adjacent} }}$

For csc, sec and cot - they are reciprocal of sin,cos and tan.

$\tt{ \large{csc \theta = \frac{1}{sin \theta} } }\\ \tt{ \large{sec \theta = \frac{1}{cos \theta} }} \\ \tt{ \large{cot \theta = \frac{1}{tan \theta} }}$

What we know now is our hypotenuse, adjacent and opposite length.

• hypotenuse = 15
• opposite = 12
• adjacent = 9

Therefore,

$\large{sin \theta = \frac{12}{15} \longrightarrow \frac{4}{5} } \\ \large{cos \theta = \frac{9}{15} \longrightarrow \frac{3}{5} } \\ \large{ tan \theta = \frac{12}{9} \longrightarrow \frac{4}{3} }$As for the reciprocal of three trigonometric ratio. We just swap the numerator and denominator.

$\large{csc \theta = \frac{15}{12} \longrightarrow \frac{5}{4} } \\ \large{sec \theta = \frac{15}{9} \longrightarrow \frac{5}{3} } \\ \large{cot \theta = \frac{9}{12} \longrightarrow \frac{3}{4} }$

Answer

• sin = 12/15 —> 4/5
• cos = 9/15 —> 3/5
• tan = 12/9 —> 4/3
• csc = 15/12 —> 5/4
• sec = 15/9 —> 5/3
• cot = 9/12 —> 3/4

The first is non-simplifed form while the second that has the arrow pointing is the simplest form.

Let me know if you have any doubts.