This chapter is currently out of syllabus
1
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Out of Syllabus
The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3$$. If c = 4 cm, then the area (in sq. cm) of this triangle is :
A
2$$\sqrt 3$$
B
4$$\sqrt 3$$
C
$${4 \over {\sqrt 3 }}$$
D
$${2 \over {\sqrt 3 }}$$
2
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Out of Syllabus
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :
A
5 : 9 : 13
B
5 : 6 : 7
C
4 : 5 : 6
D
3 : 4 : 5
3
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
Out of Syllabus
Given $${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$$ for a $$\Delta$$ABC with usual notation.

If   $${{\cos A} \over \alpha } = {{\cos B} \over \beta } = {{\cos C} \over \gamma },$$ then the ordered triad ($$\alpha$$, $$\beta$$, $$\gamma$$) has a value :
A
(19, 7, 25)
B
(7, 19, 25)
C
(5, 12, 13)
D
(3, 4, 5)
4
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
Out of Syllabus
In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 – c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :
A
$${y \over {\sqrt 3 }}$$
B
$${c \over 3}$$
C
$${c \over {\sqrt 3 }}$$
D
$${3 \over 2}$$y
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