solve using law of sines or cosines A=?,B=51°,C=?,a=6m,b=?,c=7m
cosines will work well here, since you have B,a,c
b^2 = a^2+c^2-2ac*cosB
Then you can use sines to get A,C,b
To solve this triangle using the law of sines or cosines, we need to have either A or C, as well as the lengths of two sides. However, in this case, we are given angle B and the lengths of sides a and c.
Let's start by using the law of sines, which states that the ratio of the length of a side to the sine of its opposite angle is the same for all sides of a triangle.
Let's find angle A first.
Using the law of sines, we can write the equation as:
sin(A) / a = sin(B) / b
Substituting the known values:
sin(A) / 6 = sin(51°) / b
Now, let's isolate sin(A):
sin(A) = (6 / b) * sin(51°)
To find A, we need to take the inverse sine (sin^-1) of both sides:
A = sin^-1((6 / b) * sin(51°))
Next, let's find angle C. We know that the sum of angles in a triangle is always 180°:
A + B + C = 180°
Substituting the given values:
A + 51° + C = 180°
Now, we can solve for angle C:
C = 180° - A - 51°
Once we have angle C, we can find side b using the law of sines:
sin(C) / c = sin(B) / b
Substituting the known values:
sin(C) / 7 = sin(51°) / b
Now, let's isolate b:
b = (7 / sin(C)) * sin(51°)
Finally, we have found angle A, angle C, and side b using the law of sines.