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There exists a triangle ABC satisfying the conditions
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
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 d_{1}, d_{2}, d_{3} be the altitudes on QR, RP and PQ respectively.
⇒
⇒ d_{1}, d_{2}, d_{3} are in H.P. (As given that sin P, sin Q, sin R are in A.P.)
Let A_{0}A_{1}A_{2}A_{3}A_{4}A_{5} be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A_{0}A_{1},A_{0}A_{2} and A_{0}A_{4} is
Given that A_{0}A_{1}A_{2}A_{3}A_{4}A_{5} is a regular hexagon inscribed in a circle of radius 1.
∴ A_{0}A_{1} = 1 = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{0}
∠A_{0}A_{1}A_{2} = 60° + 60° = 120°
In ΔA_{0}A_{1}A_{2}, 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 = x^{2} + x + 1, b = x^{2} – 1 and c = 2x + 1 is (are)
Given that a = x^{2} + x + 1, b = x^{2} – 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
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