let's recall that in the I Quadrant, x,y or cosine,sine are both positive, thus
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{I~Quadrant}{+4=b} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\stackrel{adjacent}{3}}[/tex]
The value of the tanθ is 4/3.
Given that,
The angle θ lies in Quadrant I.
The value of cosθ is 3/5.
We have to determine,
What is tanθ?
According to the question,
The value of tanθ is determined by using the following formula;
[tex]\rm Tan\theta = \dfrac{Perpendicular}{Base}[/tex]
The value of cosθ is 3/5.
On comparing with the following formula;
[tex]\rm Cos\theta = \dfrac{Base}{Hypotenuse}[/tex]
The value of the base is 3 and the hypotenuse is 5.
Then,
By using Pythagoras theorem the value of the perpendicular is,
[tex]\rm (Hypotenuse)^2=(Perpendicular)^2 +(Base)^2\\\\ (5)^2=(Perpendicular)^2 +(3)^2\\\\ 25=(Perpendicular)^2 +16\\\\ 25-9=(Perpendicular)^2\\\\ (Perpendicular)^2 = 16\\\\Perpendicular=\sqrt{16}\\\\Perpendicular=4[/tex]
Therefore,
The value of the tanθ is,
[tex]\rm Tan\theta = \dfrac{4}3}[/tex]
Hence, The value of the tanθ is 4/3.
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