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Prove that the area of a rhombus on the hypotenuse of a right-angled triangle, with one of the angles as 60o, is equal to the sum of the areas of rhombuses with one of their angles as 60o drawn on the other two sides.
Ans: Hint: Area of Rhombus of side a & one angle of 60o
= √3/2 x a x a = √3/2 a2
1/sec A+tan A =1-sin A /cos A
y=X^2/3(2X-X^2)
Trig function
l+bx2= 5000+100x2
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