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Grade: 12th pass
        
​​​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
one month ago

Answers : (1)

Samyak Jain
327 Points
							
We can calculate values of tan\pi/12 , tan\pi/8 and tan\pi/10 as
\pi/12 = 15o = (1/2) 30o, \pi/8 = (1/2) 45o and using formula tan2\ \theta = 2 tan\ \theta / (1 – tan2\ \theta).
So, value of  tan\pi/12 = 2 – \sqrt{3} = \sqrt{4} - \sqrt{3}  and  tan\pi/8 = \sqrt{2} - \sqrt{1}.
Hence, p = 4, q = 3, r = 2, s = 1.
Now, we shall calculate sin18o. Let 18o = \ \theta, then 90o = 5\ \theta
2\ \theta = 90o – 3\ \theta. Take sine on both sides.
sin2\ \theta = sin(90o – 3\ \theta) = cos3\ \theta  \Rightarrow  2sin\ \theta cos\ \theta = 4cos3\ \theta – 3cos\ \theta.
We can cancel cos\ \theta from both sides as it is not zero and write cos2\ \theta as 1 – sin2\ \theta.
2sin\ \theta = 4cos2\ \theta – 3  or  4sin2\ \theta +2sin\ \theta – 1 = 0.
Solving this quadratic we get (take only positive root as sin18o is positive)
sin18o = (\sqrt{5} - 1) / 4 . 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.
29 days ago
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