Question

# In $$\triangle ABC$$, right angled at $$B,AB=24cm,BC=7cm$$. Determine$$\sin A, \cos A$$

Solution

## $$\sin A$$we know that$$\sin { \theta } =\cfrac { side\quad opposite\quad to\quad angle\quad \theta }{ hypotenuse }$$here $$\theta =A$$side opposite to $$\angle A=BC=7$$Hypotenuse $$=AC=$$?We have to find value of $$AC$$ with the help of Pythagoras theoremAccording to Pythagoras theorem$${ \left( Hypotenuse \right) }^{ 2 }={ \left( Base \right) }^{ 2 }+{ \left( Perpendicular \right) }^{ 2 }$$$$\Rightarrow { \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( AC \right) }^{ 2 }\Rightarrow { \left( 24 \right) }^{ 2 }+{ \left( 7 \right) }^{ 2 }={ \left( AC \right) }^{ 2 }\Rightarrow 576+49={ \left( AC \right) }^{ 2 }\quad$$$$\Rightarrow { \left( AC \right) }^{ 2 }=625\Rightarrow AC=\sqrt { 625 } =\pm 25$$But side $$AC$$ can't be negative, so $$AC=25cm$$Now, $$BC=7$$ and $$AC=25$$So, $$\cos A$$we know that$$\cos { \theta } =\cfrac { side\quad adjacent\quad to\quad angle\quad \theta }{ hypotenuse }$$here $$\theta =A$$side adjacent to $$\angle A=AB=24$$Hypotenuse $$=AC=25$$So, $$\cos { A } =\cfrac { AB }{ AC } =\cfrac { 24 }{ 25 }$$And, $$\sin A=\cfrac {BC}{AC}=\cfrac{7}{25}$$Mathematics

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