In a triangle ABC a:b:c =4:5:6 then 3A+B=?
Hmmm. If the sides are in the ratio 4:5:6, then the sines of the angles are in the same ratio (law of sines)
So, if sinA = 4x,
sinB = 5x
sinC = 6x
arcsin(4x)+arcsin(5x)+arcsin(6x) = pi
x = √7/16
A = arcsin(4√7/16) = 41.41
B = arcsin(5√7/16) = 55.77
C = arcsin(6√7/16) = 82.82
In degrees, then
3A+B = 180
Wow! Unexpected. Haven't figured out another way to get that, just yet.
To find the value of 3A + B, we first need to determine the measure of angle A and angle B in triangle ABC.
Given that the ratio of the sides of the triangle ABC is a:b:c = 4:5:6, we can assume that the corresponding angles are in the same ratio. Let's assume the measure of angle A is 4x, angle B is 5x, and angle C is 6x.
Since the sum of the angles in any triangle is always 180 degrees, we can write the equation:
4x + 5x + 6x = 180
Simplifying the equation:
15x = 180
Dividing both sides by 15:
x = 12
Now we can substitute the value of x back into the expressions for angle A and angle B:
Angle A = 4x = 4 * 12 = 48 degrees
Angle B = 5x = 5 * 12 = 60 degrees
Finally, we can calculate 3A + B:
3A + B = 3 * 48 + 60 = 144 + 60 = 204 degrees
Therefore, 3A + B is equal to 204 degrees.
saurav Kumar sarwan bazar
3A+B=2A+180-C
Now sin(2A)=sin(C)=6x
So, 2A=C
3A+B=180