Plz Helppp meeee assapppp

Asked 9 months ago
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Plz Helppp meeee assapppp

1 Answer

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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}.

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Miss Amely Parisian
15.5k 3 10 26
answered 9 months ago