Given `b = 2, c = sqrt3, angle A = 30^(@)`, then inradius of `DeltaABC` is

Correct Answer - a Both the lines pass through the origin. Line `L_(1)` is parallel to the vector `vec(V_1)` `" "vec(V_1)= (costheta+sqrt(3))hati+ (sqrt2 sin theta)hatj + (costheta-sqrt3)hatk` and `L_(2)` is parallel...

Correct Answer - B Eliminating from the given two equations, we have `(x^(2))/(16)-(y^(2))/(48)=1` whose eccentricity is `e=sqrt(1+(48)/(16))=2`

Correct Answer - A Given are centroid `G(-(2)/(sqrt3),(2)/(sqrt3))` and one vertex `A(2,2)` `therefore` Length of side of triangle `=sqrt(3)GA` `=sqrt(3)sqrt((-(2)/(sqrt3)-2)^2+((2)/(sqrt3)-2)^2)` `=2sqrt((1+sqrt3)^2+(1-sqrt3)^2)=2sqrt8=4sqrt2`

Correct Answer - B `(1)/(2) bc sin.(2pi)/(3) = (9sqrt3)/(2)` `rArr bc = 18` Now, we have `b - c = 3sqrt3` `rArr b^(2) + c^(2) - 36 = 27` [using (1)]...

Correct Answer - A Since `DeltaABC` is right angled at C, circum-radius, `R = (c)/(2)` Now, `r = (s -c) tan (C//2) = (s-c) tan (pi//4) = s -c` Thus, `2(r...

Correct Answer - B Let a and b the roots of `x^(2) -7 x + 8 = 0` Then `a + b = 7, ab = 8` Also, `C = 60^(@)`...

Correct Answer - D We have `r_(1) -r 4R sin.(A)/(2) [cos.(B)/(2) cos.(C)/(2) - sin.(B)/(2) sin.(C)/(2)]` `rArr 7 -1 = 4R sin.(A)/(2) cos ((B+C)/(2))` `rArr 6 = 4R sin^(2).(A)/(2)` `rArr 6 =...

Correct Answer - A::B `c^(2) = a^(2) + b^(2) - 2ab cos C` `=(a - b)^(2) + 2ab (1-cos C)` `=(a -b)^(2) + (4Delta)/(sinC) (1-cosC)` For c to be minium, `a...

Correct Answer - A::C `|BD -CD| = |(s-b) - (s-c)|` `= |b-c|` `= 2R sin.(B-C)/(2)."cos"(B+C)/(2)` `=|4R sin.(B-C)/(2)."sin"(A)/(2)|`

Correct Answer - D `(Delta)/(s-a) : (Delta)/(s-b): (Delta)/(s-c) = 1: 2:3` Let `(Delta)/((s-a)/(1)) = (Delta)/((s-b)/(2)) = (Delta)/((s-c)/(3)) = (Delta)/(6k)` `rArr (1)/(s-a) = (1)/(6k), (1)/(s-b)= (1)/(3k), (1)/(s-c) = (1)/(2k)` `rArr s-a =...