Sin[tex]\theta_{1}=\frac{3\sqrt{21}}{17}[/tex]
Solution given:
Cos[tex]\theta_{1}=\frac{10}{17}[/tex]
[tex]\frac{adjacent}{hypotenuse}=\frac{10}{17}[/tex]
equating corresponding value
we get
adjacent=10
hypotenuse=17
perpendicular=x
now
by using Pythagoras law
Hypotenuse ²=perpendicular²+adjacent ²
substituting value
17²=x²+10²
17²-10²=x²
x²=17²-10²
x²=189
doing square root
[tex]\sqrt{x²}=\sqrt{189}[/tex]
x=[tex]3\sqrt{21}[/tex]
now
In I Quadrant sin angle is positive
Sin[tex]\theta_{1}=\frac{perpendicular}{hypotenuse}[/tex]
Sin[tex]\theta_{1}=\frac{3\sqrt{21}}{17}[/tex]
