Evaluate `∫ 2^x/ sqrt(1 − 4^x ) d x `

Answer

`a rarr p; b rarr q, s; c rarr p; d rarr p,q`

Explanation

`sin A , sin B` are roots of `c^(2) x^(2) - c (a +b) x + ab = 0`
`:. sin A + sin B = (c(a+b))/(c^(2))`
`rArr ((a+b))/(2R) = (c(a+b))/(c^(2)) " or " 2R = c` ...(i)
From sine formula, we have
`(a)/(sin A) = (b)/(sin B) = (c)/(sin C) = 2R = c` [using Eq. (i)] ltbr or `sin C = 1 " or " C = 90^(@)`
b. Using cosine formula, we have
`cos 30^(@) = (40^(2) + (40 sqrt3)^(2) - a^(2))/(2 xx 40 xx 40 sqrt3)`
or `a = 40`

Thus, `AB = BC = 40`, i.e., `DeltaABC` is isoceles.
Also, `angleA = angleC = 30^(@)`
`rArr angle B = 120^(@)`
Therefore, `DeltaABC` is an obtuse -angled triangle
c. `81^(sin^(2)x) + 81^(cos^(2)x) = 30`
or `81^(sin^(2)x) + (81)/(81^(sin^(2)x)) = 30`
Put `81^(sin^(2)x) = t`
`t + (81)/(t) = 30`
or `t = 27, t = 3`
When `t = 3, 81^(sin^(2)x) = 3`
or `3^(4 sin^(2)x) = 3`
or `x = 30^(@)`
When `t = 27, 3^(4 sin^(2)x) = 3^(3)`
`rArr x = 60^(@)`
Therefore, the triangle is right angled
d. In `DeltaABC`,
`cos A cos B + sin A sin B sin C = 1`
`rArr sin C = (1 - cos A cos B)/(sin A sin B)`...(i)
[ `:' angleC` is an angle of the triangle]
Since `0 lt sin C le 1`, we have
`0 lt (1-cos A cos B)/(sin A sin B) le 1`
or `1- cos A cos B le sin A sin B`
or `1 le sin A sin B + cos A cos B`
or `1 le cos (A -B) or cos (A- B) ge 1`
or `cos(A-B) =1`
or `A - B = 0 " or " A = B` ....(ii)
Also from Eq. (i) and (ii),
`sinC = (1- cos^(2)A)/(sin^(2)A)`
`sin C = 1 " or " C = 90^(@)`....(iii)
From Eqs. (ii) and (iii), `DeltaABC` is a right -angled isosceles triangle