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# In a right angle triangle ABC right angled at B Find all ratios.sin⁡A=

Solution

## We have,$$\sin A =$$ $$\dfrac{Perpendicular}{Hypotenuse} \, = \, \dfrac{3}{5}$$ So, we draw a right triangle right angled at B such thatPerpendicular $$= BC = 3$$ unit and, Hypotenuse $$= AC = 5$$ units.By Pythagoras theorem, we have$$AC^2 \, = AB^2 \, + \, BC^2$$$$\Rightarrow \, \, \, \, 5^2 \, = \, AB^2 \, + \, 3^2$$$$\Rightarrow \, \, \, \, AB^2 \, = \, 25 \, - \, 9 \, = \, 16$$ $$\Rightarrow \, \, \, \, AB \, = \, \sqrt{16} \, = \, 4$$When we consider the trigonometric ratios of $$\angle$$ C , we haveBase $$= BC = 3$$, Perpendicular $$= AB = 4$$, and Hypotenuse $$= AC = 4$$$$\therefore \, \, \, \, \sin \, C \, = \, \dfrac{Perpendicular}{Hypotenuse} \, = \, \dfrac{4}{5}$$$$\cos \, C \, = \, \dfrac{Base}{Hypotenuse} \, = \, \dfrac{3}{5}$$$$\tan \, C \, = \, \dfrac{Perpendicular}{Base} \, = \, \dfrac{4}{3}$$$$\text{cosec} \, C \, = \, \dfrac{Hypotenuse}{Perpendicular} \, = \, \dfrac{5}{4}$$$$\sec \, C \, = \, \dfrac{Hypotenuse}{Base} \, = \, \dfrac{5}{3}$$and, $$\cot \, C \, = \, \dfrac{Base}{Perpendicular} \, = \, \dfrac{3}{4}$$Mathematics

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