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1) We have here a right triangle, so we can use the Pythagorean Theorem to find the adjacent leg. Note that the hypotenuse is leg opposite to the right angle. In this case, a=4.
a²=b²+c²
4²=b²+3²
16=b²+9
16-9=b²
b=√7
2) Let's find the trigonometric functions. The first one is the sine (θ), which relates the opposite side and the hypotenuse:
[tex]\sin (\theta)\text{ =}\frac{3}{4}[/tex]
2.2)Moving on, we can deal with the tangent of theta, relating the opposite side over the adjacent, for this one we have to rationalize it:
[tex]\begin{gathered} \tan (\theta)\text{ =}\frac{3}{\sqrt[]{7}} \\ \tan (\theta)=\frac{3\sqrt[]{7}}{7} \end{gathered}[/tex]
Finally, let's find the reciprocal function of the cosine function by setting this as the reciprocal of the cosine, and then calculating it:
[tex]\begin{gathered} \sec \text{ (}\theta)=\frac{1}{\cos(\theta)}\Rightarrow\sec (\theta)=\frac{1}{\frac{\sqrt[]{7}}{4}}\Rightarrow\sec (\theta)\text{ =}\frac{4}{\sqrt[]{7}} \\ \sec (\theta)\text{ =}\frac{4\sqrt[]{7}}{7} \end{gathered}[/tex]
