Consider the triangle ABC. Suppose that C=69 degrees, a=45 and b= 9 Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. List all solutions
A=
B=
c=
by law of cosines,
c^2 = a^2 + b^2 - 2ab cosC
Now, having a,b,c and C, use the law of since to get A,B.
sinA/a = sinB/b = sinC/c
To find the missing sides and angles of triangle ABC given the information provided, we can use the Law of Sines and the fact that the sum of the interior angles of a triangle is 180 degrees.
1. Use the Law of Sines to find angle A:
sin(A) / a = sin(C) / c
sin(A) / 45 = sin(69) / c
sin(A) = (45 * sin(69)) / c
A = arcsin((45 * sin(69)) / c)
2. Use the Law of Sines to find angle B:
sin(B) / b = sin(C) / c
sin(B) / 9 = sin(69) / c
sin(B) = (9 * sin(69)) / c
B = arcsin((9 * sin(69)) / c)
3. Use the fact that the sum of the angles of a triangle is 180 degrees to find angle A:
A = 180 - C - B
Now let's substitute the given values and calculate the solutions:
1. Calculate A:
A = arcsin((45 * sin(69)) / c)
A = arcsin((45 * sin(69)) / 9)
A ≈ 59.2 degrees
2. Calculate B:
B = arcsin((9 * sin(69)) / c)
B = arcsin((9 * sin(69)) / 9)
B ≈ 51.8 degrees
3. Calculate c:
We can use the Law of Sines with angle A or B to find c:
sin(A) / a = sin(C) / c
sin(59.2) / 45 = sin(69) / c
c = (45 * sin(69)) / sin(59.2)
c ≈ 52.9
or
sin(B) / b = sin(C) / c
sin(51.8) / 9 = sin(69) / c
c = (9 * sin(69)) / sin(51.8)
c ≈ 9.9