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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 theorem
According 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, $$\sin A=\cfrac {BC}{AC}=\cfrac{7}{25}$$

1790745_1813946_ans_3b800d6998aa40c8a4b88c8bd923a74e.JPG

Mathematics

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