Test: MCQs (One or More Correct Option): Properties of Triangle | JEE Advanced - JEE MCQ

We have

(a) b sin A = a ⇒ a sin B = a
⇒ sin B = 1  ⇒ B = π/2
Since A < π/2, the ΔABC is possible.
(b) b sin A > a ⇒ a sin B > a ⇒ sin B > 1
Which is impossible. Hence the πossibility (b) is ruled out.
Similarly (c) can be shown to be impossible.
(d) b sin A < a ⇒ a sin B < a ⇒ sin B < 1 so value of ∠B exists.
Now, b > a ⇒ B > A.  Since A < π/2
The ΔABC is possible when B > or < π/2.
(e) Since b = a, we have B = A. But A > π/2
∴ B > π/2. But this is not possible for any triangle

Since the angles of triangle are in A.P., Let ∠A = x – d,  ∠B = x,  ∠C = x + d
Then by ∠ sum property of triangle, we have ∠A +∠ B + ∠C = 180°
⇒ x  – d + x + x + d = 180°
⇒ 3x = 180° ⇒ x = 60°
∴ ∠B = 60°

Given that a = 10,  b = 9 are the longer sides
 (both the values are less than 9 and 10),
 both are possible.

In ΔPQR let d₁, d₂, d₃ be the altitudes on QR, RP and PQ respectively.

⇒ 

⇒ d₁, d₂, d₃  are  in H.P. (As given that sin P,  sin Q,  sin R  are in A.P.)

Given that A₀A₁A₂A₃A₄A₅ is a regular hexagon inscribed in a circle of radius 1.

∴ A₀A₁ = 1 = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₀ 
∠A₀A₁A₂ = 60°  +  60° = 120°
In ΔA₀A₁A₂, using cosine law we get


∴ (c) is the correct option.

By simple geometry in ΔAFE, AF = AE
∴ ΔAFE is  an isosceles Δ.
Now Ar (ΔABC) = Ar (ΔABD) + Ar (ΔADC)

Given that a = x² + x + 1, b = x² – 1, c = 2x+ 1 and ∠C = π/6

Now for x = –1and 1, b = 0 which is not possible
Also fo which is not possible

Let PN = x, QL = x + 2, RM = x + 4 where x is an even integer.

Then PM = PN = x, QN = QL = x + 2 and RL = RM = x + 4
So that PQ = 2x + 2, QR = 2x + 6, PR = 2x + 4