To find the largest angle in triangle ABC, you can use the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:
c^2 = a^2 + b^2 - 2ab * cos(C)
Where:
- c is the side opposite the angle C (the largest angle in this case).
- a and b are the lengths of the other two sides.
- C is the angle we want to find.
You know that the perimeter of the triangle is 17 cm, so the sum of all three sides must be 17 cm:
a + b + c = 17
You also know that a = 4.5 cm and b = 5 cm. Plugging these values into the equation, you can solve for c:
4.5 + 5 + c = 17
c = 17 - 4.5 - 5
c = 7.5 cm
Now that you have the lengths of all three sides, you can use the Law of Cosines to find the largest angle, C:
7.5^2 = 4.5^2 + 5^2 - 2 * 4.5 * 5 * cos(C)
Now, solve for cos(C):
cos(C) = (4.5^2 + 5^2 - 7.5^2) / (2 * 4.5 * 5)
cos(C) = (20.25 + 25 - 56.25) / (2 * 4.5 * 5)
cos(C) = (20.25 - 31) / (2 * 4.5 * 5)
cos(C) = (-10.75) / 45
cos(C) = -0.2389
Now, find the largest angle C by taking the arccosine of -0.2389:
C = arccos(-0.2389)
C ≈ 104 degrees
So, the largest angle in triangle ABC is approximately 104 degrees. Therefore, the answer is 104°.