sec⁡θ=3,sin⁡θ&gt;0 Find sin⁡θ

since sec⁡θ=3&gt;0 is positive and sin⁡θ&gt;0 then θ is in Quadrant I. We know that: sec⁡θ=hypopp=31 So (opp)2+(adj)2=(hyp)2 (opp)2+12=32 (opp)2+1=9 (opp)2=8 opp=8=22 Hence sin⁡θ=opphypp sin⁡θ=223

Explained: "sec⁡θ=3,sin⁡θ>0 Find sin⁡θ"

High School
Geometry
Trigonometry
Right triangles and trigonometry

Ava-May Nelson

Answered question

2020-12-15

\sec θ = 3, \sin θ > 0

Find

\sin θ

Answer & Explanation

gwibdaithq

Skilled2020-12-16Added 84 answers

since $\sec θ = 3 > 0$ is positive and $\sin θ > 0 then θ$ is in Quadrant I.
We know that:
$\sec θ = \frac{h y p}{o p p} = \frac{3}{1}$
So
$(o p p)^{2} + (a d j)^{2} = (h y p)^{2}$
$(o p p)^{2} + 1^{2} = 3^{2}$
$(o p p)^{2} + 1 = 9$
$(o p p)^{2} = 8$
$o p p = \sqrt{8} = \frac{2}{\sqrt{2}}$
Hence
$\sin θ = \frac{o p p}{h y p p}$
$\sin θ = 2 \frac{\sqrt{2}}{3}$

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