Let `A(0,beta), B(-2,0)` and `C(1,1)` be the vertices of a triangle. Then All the values of `beta` for which angle A of triangle ABC is largest lie in the intervalA. `(-2,1)`B. `(-2,(2)/(3))uu((2)/(3),1)`C. `(-2,(2)/(3))uu((2)/(3),sqrt(6))`D. none of these

Let `A(0,beta), B(-2,0)` and `C(1,1)` be the vertices of a triangle. T

1.	Let `A(0,beta), B(-2,0)` and `C(1,1)` be the vertices of a triangle. Then All the values of `beta` for which angle A of triangle ABC is largest lie in the intervalA. `(-2,1)`B. `(-2,(2)/(3))uu((2)/(3),1)`C. `(-2,(2)/(3))uu((2)/(3),sqrt(6))`D. none of these
Answer» Correct Answer - C Whenever A is obtuse is the largest angle also. But there many be more values of `beta` for which A is largest but is not obtuse. Angle A will be largest if `BC gt AC, BC gt AB, A,B,C` are not collinear `rArr 10 gt 1 +(beta -1)^(2),10 gt 4 +beta^(2), beta ne (2)/(3)` `rArr beta^(2)-2beta -8 lt 0, beta^(2) lt 6` `rArr-2 lt beta lt 4, -sqrt(6) lt beta sqrt(6)` `beta in (-2,(2)/(3))uu((2)/(3),sqrt(6))`

If `A(5,2),B(10,12)` and `P(x,y)` is such that `(AP)/(PB) = (3)/(2)`, then then internal bisector of `/_APB` always passes throughA. `(20,32)`B. `(8,8)`C. `(8,-8)`D. `(-8,-8)`
Let `A(0,beta), B(-2,0)` and `C(1,1)` be the vertices of a triangle. Then All the values of `beta` for which angle A of triangle ABC is largest lie in the intervalA. `(-2,1)`B. `(-2,(2)/(3))uu((2)/(3),1)`C. `(-2,(2)/(3))uu((2)/(3),sqrt(6))`D. none of these
`A(x_1,y_1), B(x_2,y_2),C(x_3,y_3)` are three vertices of a triangle ABC, `lx+my+n=0` is an equation of line L. If L intersects the sides BC,CA and AB of a triangle ABC at P,Q,R respectively, then `(BP)/(PC)xx(CQ)/(QA)xx(AR)/(RB)` is equal toA. `-1`B. `-(1)/(2)`C. `(1)/(2)`D. 1
Number of values of `alpha` such that the points `(alpha,6),(-5,0)` and (5,0) form an isosceles triangle isA. 4B. 5C. 6D. 7
The maximum value of `y = sqrt((x-3)^(2)+(x^(2)-2)^(2))-sqrt(x^(2)-(x^(2)-1)^(2))` isA. 3B. `sqrt(10)`C. `2sqrt(5)`D. none of these
If `m_(1),m_(2)` be the roots of the equation `x^(2)+(sqrt(3)+2)x+sqrt(3)-1 =0`, then the area of the triangle formed by the lines `y = m_(1)x,y = m_(2)x` and `y = 2` isA. `sqrt(33)-sqrt(11)` sq. unitsB. `sqrt(11) +sqrt(33)` sq. unitsC. `2sqrt(33)` sq. unitsD. 121 sq. units