Find the size of angle C if angle A is 50 degrees. You have been give that triangle ABC AC = 8 cm and AB = 6 cm.
50+C=180 C=180-50 C=130
Nope. The A+B+C=180
Note that the sides
AC = b
AB = c
Use the law of cosines to get
a^2 = b^2+c^2-2bc cosA
Now, having a, you can use the law of sines to get A:
sinA/a = sinC/c
To find the size of angle C, we can use the law of cosines.
The law of cosines is a formula used to find the length of a side or the measure of an angle in a triangle when we know the lengths of the other sides and/or the measures of the other angles.
In this case, we know the lengths of sides AC and AB, and we want to find the size of angle C. We can use the law of cosines as follows:
c^2 = a^2 + b^2 - 2ab*cos(C)
where c is the side opposite angle C, a is the side opposite angle A, and b is the side opposite angle B.
Let's substitute the given values into the formula:
AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(C)
Since we know that AC = 8 cm and AB = 6 cm, we can rewrite the equation as:
(8)^2 = (6)^2 + BC^2 - 2*(6)*(BC)*cos(C)
Simplifying further:
64 = 36 + BC^2 - 12BC*cos(C)
Rearranging the equation:
BC^2 - 12BC*cos(C) = 64 - 36
BC^2 - 12BC*cos(C) = 28
Now, we need more information to proceed. Do you have any other angles or side lengths given for triangle ABC?