If the angles of A, B and C of a ΔABC are in A.P. and the sides a, b and c opposite to these angles are in G.P., then a2, b2 and c2 are in –
Question
If the angles of A, B and C of a ΔABC are in A.P. and the sides a, b and c opposite to these angles are in G.P., then a², b² and c² are in –
Solution
The given conditions are that the angles A, B, and C of a triangle ABC are in Arithmetic Progression (A.P.) and the sides a, b, and c opposite to these angles are in Geometric Progression (G.P.).
From the given conditions, we can write:
B = A + d and C = A + 2d (since the angles are in A.P.)
Also, b = ar and c = ar^2 (since the sides are in G.P.)
Now, we know that in any triangle, by the law of cosines, we have:
c^2 = a^2 + b^2 - 2ab cosC b^2 = a^2 + c^2 - 2ac cosB a^2 = b^2 + c^2 - 2bc cosA
Substituting the values of B and C from above in these equations, we get:
c^2 = a^2 + b^2 - 2ab cos(A + 2d) b^2 = a^2 + c^2 - 2ac cos(A + d) a^2 = b^2 + c^2 - 2bc cosA
Solving these equations, we find that a^2, b^2, and c^2 are in Arithmetic Progression (A.P.).
Similar Questions
If in ∆ ABC and ∆ PQR, ொோ = ோ = ொ then:a) ∆PQR ~∆CAB b) ∆PQR ~∆ABC c) ∆CBA ~∆PQR d) ∆BCA ~∆PQR
In the given figure, equilateral ΔABD and ΔACE are drawn on the sides of a ΔABC. Prove that CD = BE.
In a ΔABC, P and Q are points on AB and AC respectively, such that AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC = 3 PQ.TrueFalse
300. Let A = (5, −3, 6), B = (0, 0, 0), and C = (3, 7, 1). Show that ABC is a right angle
ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle.Explain why O is equidistant from A, B, and C.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.