Consider A C B right angled at C in which AB =29 units, BC =21 un… - Vidyadip

Question

Consider $\triangle A C B$ right angled at C in which $AB =29$ units, $BC =21$ units and $\angle A B C=\theta$. Determine the values of
i. $\cos ^2 \theta+\sin ^2 \theta$
ii. $\cos ^2 \theta-\sin ^2 \theta$

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Answer

In, $ \Delta A C B$ we have
$ A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }$
$ \Rightarrow \quad A C = \sqrt { A B ^ { 2 } - B C ^ { 2 } } = \sqrt { 29 ^ { 2 } - 21 ^ { 2 } } = \sqrt { ( 29 + 21 ) ( 29 - 21 ) } = \sqrt { 400 } = 20$ units
$ \therefore \quad \sin \theta = \frac { A C } { A B } = \frac { 20 } { 29 }$and $ \cos \theta = \frac { B C } { A B } = \frac { 21 } { 29 }$

Using the values of $ \sin \theta$ and,$ \cos \theta$ we get
$ \cos ^ { 2 } \theta + \sin ^ { 2 } \theta = \left( \frac { 21 } { 29 } \right) ^ { 2 } + \left( \frac { 20 } { 29 } \right) ^ { 2 }$
$ = \frac { 441 + 400 } { 841 } = 1$
Using the values of $ \sin \theta$and,$ \cos \theta$ we obtain
$\cos^2\theta-\sin^2\theta=\left(\frac{21}{29}\right)^2-\left(\frac{20}{29}\right)^2=\frac{21^2-20^2}{29^2}=\frac{(21+20)(21-20)}{841}=\frac{41}{841}$

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