Question

Plz Helppp meeee assapppp

Answer 1

Given:

A right angled triangle.

To find:

The value of x.

Solution:

We have,

Base = x

Hypotenuse = $\sqrt{5}$

In a right angle triangle,

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

For the given triangle,

$\cos 45^\circ= \dfrac{x}{\sqrt{5}}$

$\dfrac{1}{\sqrt{2}}= \dfrac{x}{\sqrt{5}}$

Multiply both sides by $\sqrt{5}$.

$\dfrac{\sqrt{5}}{\sqrt{2}}= x$

$\dfrac{\sqrt{5}}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}= x$

$\dfrac{\sqrt{10}}{2}= x$

Therefore, the required value of x is $\dfrac{\sqrt{10}}{2}$.