Question

In a right angled triangke abc the perpendicular and the base measure 3 cm find hypotenus and three trignometrice ratios

Answer 1

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:

$Hypotenuse^2=Perpendicular^2+Base^2$

Use Pythagoras theorem in triangle ABC.

$Hypotenuse^2=(3)^2+(3)^2$

$Hypotenuse^2=9+9$

$Hypotenuse=\sqrt{18}$

$Hypotenuse=3\sqrt{2}$

The three trigonometric ratios are

$\sin \theta=\dfrac{Perpendicular}{Hypotenuse}$

$\sin \theta=\dfrac{3}{3\sqrt{2}}$

$\sin \theta=\dfrac{1}{\sqrt{2}}$

Similarly,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{3}{3\sqrt{2}}$

$\cos \theta=\dfrac{1}{\sqrt{2}}$

And,

$\tan \theta=\dfrac{Perpendicular}{Base}$

$\tan \theta=\dfrac{3}{3}$

$\tan \theta=1$

Therefore, the hypotenuse is $3\sqrt{2}$ cm and three trigonometric ratios are $\sin \theta=\dfrac{1}{\sqrt{2}}, \cos \theta=\dfrac{1}{\sqrt{2}}, \tan \theta=1$.