If the sides of ∆ABC are 4,5 6 then prove that largest angle will be twice of smallest angle
To prove that the largest angle in triangle ABC will be twice the measure of the smallest angle, we can use the concept of the Law of Cosines.
The Law of Cosines states that in any triangle with sides a, b, and c, and the opposite angles A, B, and C respectively, the following relationship holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
Let's assign the lengths of the sides of triangle ABC as follows:
a = 4
b = 5
c = 6
To find the measures of the angles, let's denote them as angles A, B, and C, where A is the largest angle, and C is the smallest angle.
According to the Law of Cosines, we have:
c^2 = a^2 + b^2 - 2ab * cos(C)
Substituting the given values, we get:
6^2 = 4^2 + 5^2 - 2(4)(5) * cos(C)
36 = 16 + 25 - 40 * cos(C)
36 = 41 - 40 * cos(C)
40 * cos(C) = 41 - 36
40 * cos(C) = 5
cos(C) = 5/40
cos(C) = 1/8
Now, we can use the inverse cosine (arccos) function to find the measure of angle C:
C = arccos(1/8)
Using a scientific calculator, we find:
C ≈ 82.87°
Since angle A is the largest angle, it will be twice the measure of the smallest angle C:
A = 2 * C
A = 2 * 82.87°
A ≈ 165.74°
Therefore, we have proven that the largest angle in triangle ABC is twice the measure of the smallest angle.
6^2 = 4^2 + 5^2 - 2(4)(5)cosØ
cosØ = (16+25-36)/(40) = 1/8
Ø = 82.819..
4^2 = 5^2 + 6^2 - 2(5)(6)cos A
cosA = (25+36-16)/60 = 3/4
A = 41.4096..
and 2(41.4096..) = 82.81924... same as Ø
or ... an actual "proof"
from above:
cos(smallest angle) = cosA = 3/4
cos(largest angle) = 1/8
we have to show that cos 2A = 1/8
cos 2A = 2cos^2 A - 1 , one of our main identities
= 2(9/16) - 1
= 9/8 - 1
= 1/8