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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
6 months ago

## Answers : (1)

Samyak Jain
333 Points
							We can calculate values of tan$\dpi{100} \pi$/12 , tan$\dpi{100} \pi$/8 and tan$\dpi{100} \pi$/10 as$\dpi{100} \pi$/12 = 15o = (1/2) 30o, $\dpi{100} \pi$/8 = (1/2) 45o and using formula tan2$\dpi{100} \ \theta$ = 2 tan$\dpi{100} \ \theta$ / (1 – tan2$\dpi{100} \ \theta$).So, value of  tan$\dpi{100} \pi$/12 = 2 – $\dpi{80} \sqrt{3}$ = $\dpi{80} \sqrt{4} - \sqrt{3}$  and  tan$\dpi{100} \pi$/8 = $\dpi{80} \sqrt{2} - \sqrt{1}$.Hence, p = 4, q = 3, r = 2, s = 1.Now, we shall calculate sin18o. Let 18o = $\dpi{100} \ \theta$, then 90o = 5$\dpi{100} \ \theta$2$\dpi{100} \ \theta$ = 90o – 3$\dpi{100} \ \theta$. Take sine on both sides.sin2$\dpi{100} \ \theta$ = sin(90o – 3$\dpi{100} \ \theta$) = cos3$\dpi{100} \ \theta$  $\dpi{100} \Rightarrow$  2sin$\dpi{100} \ \theta$ cos$\dpi{100} \ \theta$ = 4cos3$\dpi{100} \ \theta$ – 3cos$\dpi{100} \ \theta$.We can cancel cos$\dpi{100} \ \theta$ from both sides as it is not zero and write cos2$\dpi{100} \ \theta$ as 1 – sin2$\dpi{100} \ \theta$.2sin$\dpi{100} \ \theta$ = 4cos2$\dpi{100} \ \theta$ – 3  or  4sin2$\dpi{100} \ \theta$ +2sin$\dpi{100} \ \theta$ – 1 = 0.Solving this quadratic we get (take only positive root as sin18o is positive)sin18o = $\dpi{80} (\sqrt{5} - 1) / 4$ . 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.

6 months ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions