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        ​​​tanπ\12=√p--√q , tanπ\8=√r-√s and sinπ\10=√t-√s/p where p,q,r,s,t€N. then find the ratio of the area of triangle formed by side whose length are equal to p,q and t and rectangle formed by adjacent side of length r and s
10 months ago

							We can calculate values of tan/12 , tan/8 and tan/10 as/12 = 15o = (1/2) 30o, /8 = (1/2) 45o and using formula tan2 = 2 tan / (1 – tan2).So, value of  tan/12 = 2 –  =   and  tan/8 = .Hence, p = 4, q = 3, r = 2, s = 1.Now, we shall calculate sin18o. Let 18o = , then 90o = 52 = 90o – 3. Take sine on both sides.sin2 = sin(90o – 3) = cos3    2sin cos = 4cos3 – 3cos.We can cancel cos from both sides as it is not zero and write cos2 as 1 – sin2.2sin = 4cos2 – 3  or  4sin2 +2sin – 1 = 0.Solving this quadratic we get (take only positive root as sin18o is positive)sin18o =  . Thus, t = 5.Clearly, triangle formed by sides 3, 4 and 5 is a right angled triangle, whose adjacent sides to 90o are 3 and 4.Its area is (1/2) x 3 x 4 = 6 sq unit.Area of rectangle formed by adjacent sides 2 and 1 is 2 x 1 = 2 sq unit.Required ratio is 6 : 2  or  3 : 1.

10 months ago
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