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Question

In a right angle triangle ABC right angled at B Find all ratios.sin⁡A=
1008410_fe5bee6c86344181b80b34337dcfb936.PNG


Solution

We have,

$$\sin A =$$ $$\dfrac{Perpendicular}{Hypotenuse} \, = \, \dfrac{3}{5}$$ 
So, we draw a right triangle right angled at B such that
Perpendicular $$= BC = 3$$ 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 have
Base $$= BC = 3$$, 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, 
$$\cot \, C \, = \, \dfrac{Base}{Perpendicular} \, = \, \dfrac{3}{4}$$

1038191_1008410_ans_7965f93800df4245bfc3b8c608e5ee7b.PNG

Mathematics

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