Respuesta :
Given:
In a right angled triangle ABC the perpendicular and the base measure 3 cm.
To find:
The hypotenuse and three trigonometric ratios.
Solution:
Pythagoras theorem:
[tex]Hypotenuse^2=Perpendicular^2+Base^2[/tex]
Use Pythagoras theorem in triangle ABC.
[tex]Hypotenuse^2=(3)^2+(3)^2[/tex]
[tex]Hypotenuse^2=9+9[/tex]
[tex]Hypotenuse=\sqrt{18}[/tex]
[tex]Hypotenuse=3\sqrt{2}[/tex]
The three trigonometric ratios are
[tex]\sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin \theta=\dfrac{3}{3\sqrt{2}}[/tex]
[tex]\sin \theta=\dfrac{1}{\sqrt{2}}[/tex]
Similarly,
[tex]\cos \theta=\dfrac{Base}{Hypotenuse}[/tex]
[tex]\cos \theta=\dfrac{3}{3\sqrt{2}}[/tex]
[tex]\cos \theta=\dfrac{1}{\sqrt{2}}[/tex]
And,
[tex]\tan \theta=\dfrac{Perpendicular}{Base}[/tex]
[tex]\tan \theta=\dfrac{3}{3}[/tex]
[tex]\tan \theta=1[/tex]
Therefore, the hypotenuse is [tex]3\sqrt{2}[/tex] cm and three trigonometric ratios are [tex]\sin \theta=\dfrac{1}{\sqrt{2}}, \cos \theta=\dfrac{1}{\sqrt{2}}, \tan \theta=1[/tex].
