|1 Crore+ students have signed up on EduRev. Have you?|
There exists a triangle ABC satisfying the conditions
(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
In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A P. Then the length of the third side can be
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.
If in a triangle PQR, sin P, sin Q, sin R are in A.P., then
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.)
Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1,A0A2 and A0A4 is
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.
In ΔABC, internal angle bisector of ∠A meets side BC in D. DE ⊥ AD meets AC in E and AB in F. Then
By simple geometry in ΔAFE, AF = AE
∴ ΔAFE is an isosceles Δ.
Now Ar (ΔABC) = Ar (ΔABD) + Ar (ΔADC)
Let ABC be a triangle such that and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and c = 2x + 1 is (are)
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
In a triangle PQR, P is the largest angle and cos p = 1/3. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
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
In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X,Y, Z, respectively, and 2s = x + y + z.
and area of incircle of the triangle