Answer:
Therefore,
[tex]\tan A=\dfrac{11}{60}[/tex]
Step-by-step explanation:
Given:
[tex]\sin A=\dfrac{11}{61}[/tex]
A is in I Quadrant
To Find:
[tex]\tan A = ?[/tex]
Solution:
Using Identity
[tex]\sin^{2}A+\cos^{2}A=1[/tex]
Now Substitute Sin A we get
[tex](\dfrac{11}{61})^{2}+\cos^{2}A=1\\\\\cos^{2}A=1-\dfrac{121}{3721}=\dfrac{3600}{3721}\\\\\cos A=\pm\sqrt{\dfrac{3600}{3721}}\\\\\cos A=\dfrac{60}{61}[/tex]
As 'A' is in First Quadrant Cos A is Positive
Now Tan identity we have
[tex]\tan A=\dfrac{\sin A}{\cos A}[/tex]
Now Substitute Sin A and Cos A we get
[tex]\tan A=\dfrac{\dfrac{11}{61}}{\dfrac{60}{61}}\\\\\tan A=\dfrac{11}{60}[/tex]
Therefore,
[tex]\tan A=\dfrac{11}{60}[/tex]