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If the sides of a right-angled triangle are in AP, then the sines of the acute angles are
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a
35,45
b
3,13
c
5−12,5+12
d
3−12,3+12
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Correct option is A
Let ABC be the right triangle, right angled at B. Let the sides of the triangle be a-d, a and a + d, with a > d >0. Clearly, AC =a + d is the largest side. So, a + d is the side opposite to the right angle.By Pythagoras theorem, we have(a+d)2=a2+(a−d)2⇒4ad=a2⇒a=4dNow, sinA=BCAC=aa+d=4d5d=45and sinC=ABAC=a−da+d=4d−d4d+d=35Similar Questions
If and 1 are in A.P, then x is equal to
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