What is the angle B?
Given that
A=46 degree
b=4
c=8
Is this a triangle?
Law of Cosines:
a^2=b^2+c^2-2bcCosA
solve for a.
Then, law of sines:
sin46/a=sinB/b
solve for sinB, then B
LAW OF COSINES: CosA=(b^2+c^2-a^2)/2bc
Cos46=(16+64-a^2)/2*4*8.
Cos46=(80-a^2)/64, 80-a^2=64Cos46,
a^2=80-64Cos46=35.54, a=5.96=6.0
LAW of SINES: SinB/b=SinA/a
SinB/4=Sin46/6,Multiply both sides by 4: SinB=4*Sin46/6=0.4791, B=28.66=28.7
and easy way to check this is to grab your protractor and sketch it to see if it works.
To find the angle B, we can use the law of cosines. The law of cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides, minus twice the product of the two sides multiplied by the cosine of the included angle.
In this case, we are given side lengths b = 4 and c = 8, and the angle A = 46 degrees. We want to find angle B.
The law of cosines can be written as:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we can rearrange the formula to solve for angle B:
cos(B) = (a^2 + c^2 - b^2) / (2ac)
Let's plug in the given values:
cos(B) = (4^2 + 8^2 - 4^2) / (2*4*8)
= (16 + 64 - 16) / 64
= 64 / 64
= 1
Now, we need to find the inverse cosine (cos^-1) of 1 to get the value of angle B:
B = cos^(-1)(1)
B = 0 degrees
Therefore, angle B is 0 degrees.