Given ∆ABC. If side a is twice as long as side b, is <A necessarily twice as large as <B? Why?
no. sinA = 2 sinB
To determine whether angle A is necessarily twice as large as angle B in triangle ABC, we need to apply the principles of triangle geometry.
In a triangle, the sum of all interior angles is always 180 degrees (π radians). Let's assume that angle A is twice as large as angle B.
If we label angle A as 2x and angle B as x, we can now visualize the relationship between the angles. Additionally, we know that side a is twice as long as side b.
Using the Law of Sines, we can relate the sides and angles of a triangle. According to the Law of Sines:
sin(A)/a = sin(B)/b = sin(C)/c
In our case, let's consider side a and side b:
sin(2x)/2b = sin(x)/b
To simplify, we can cross-multiply:
2b * sin(x) = b * sin(2x)
Canceling out b from both sides, we get:
2 * sin(x) = sin(2x)
We can now use a trigonometric identity called the double angle formula:
sin(2x) = 2 * sin(x) * cos(x)
Substituting this into our equation, we have:
2 * sin(x) = 2 * sin(x) * cos(x)
Dividing both sides by 2 * sin(x), we get:
1 = cos(x)
From this equation, we can observe that x equals 60 degrees (π/3 radians). Therefore, angle B is 60 degrees (π/3 radians).
Since angle A was assumed to be twice as large as angle B, we find that angle A is 120 degrees (2 * 60 degrees) or 2π/3 radians.
To conclude, if side a is twice as long as side b in triangle ABC, angle A is necessarily twice as large as angle B when side b is opposite angle B and side a is opposite angle A.