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Find the value of sin37o,sin53o,tan37o,tan53o in terms of fraction.

Answer
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Hint: In this question, we have to write the given trigonometric function in terms of fraction.
We know, that in a right- angled triangle, there are three sides, perpendicular, base and the hypotenuse.
And, sinθ=PH , where, P is the length of perpendicular and H is referred to as the length of hypotenuse, whereas, tanθ=PB , where, P is the length of perpendicular and B is the length of base of the triangle.

Complete answer:
Given trigonometric functions sin37o,sin53o,tan37o,tan53o .
To write these trigonometric functions in terms of fraction.
Consider a right- angled triangle, ABC , with ACB=37o and ABC=90o .
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Then, we have, using angle sum property of a triangle, that, ABC+BAC+ACB=180o , i.e., 37o+90o+BAC=180o . On solving, we get, BAC=180o127o i.e., BAC=53o .
Now, let length of side AB is 3units and length of side BC is 4units , then, by Pythagoras theorem, we have, AB2+BC2=AC2 , putting values, we get, 32+42=9+16=25 , hence, AC=5units .
Now, we know, sinθ=PH , and for angle θ=37o , P=3 and H=5 . So, sin37o=35 .
Similarly, for angle θ=53o , P=4 and H=5 . So, sin53o=45 .
Now, for angle θ=37o , P=3 and B=4 . So, tan37o=34 .
Similarly, for angle θ=53o , P=4 and B=3 . So, tan53o=43

Note:
It is not necessary to choose the lengths of sides of the triangle to be 3units or 4units . We can choose the length of sides of the right- angled triangle by our choices.
If x2=a2 , then, taking square root on both sides, we get, x=±a , but in this question, we are talking about length of sides and length can never be negative. Hence, we have taken only the positive one.
For any angle, say θ , the sides opposite to this angle will be the perpendicular side, whereas, the third side except for the hypotenuse, will be the base of the triangle.