In $\triangle P Q R$, if $PQ=10\ cm$, $QR=8\ cm$ and $PR=6\ cm$ then find $\angle R=?$
Given: In $\triangle P Q R$, if $PQ=10\ cm$, $QR=8\ cm$ and $PR=6\ cm$.
To do: To find $\angle R=?$
Solution:
As given, in $\triangle P Q R$, if $PQ=10\ cm$, $QR=8\ cm$ and $PR=6\ cm$.
On squaring $PQ,\ QR$ and $PR$.
$PQ^2=10^2=100$
$QR^2=8^2=64$
And $PR^2=6^2=36$
Here we find, $PQ^2=QR^2+PR^2=36+64=100$
It is Pythagorean triplets.
Therefore, $\triangle P Q R$ is a right angled triangle at $R$.
Thus, $\angle R$ is $90^{\circ}$.
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