What is the measure to the nearest degree of the smallest angle in a triangle whose vertices are R(3, 3), S(-2, 3), and T(-2, -3)?
47º
40º
49º
51º
I calculated and got 50 but it's not an option. 49?
Nevermind, I tried a different angle and got 40.
To find the measure of the smallest angle in a triangle, we need to find the lengths of its sides or the coordinates of its vertices. In this case, we have the coordinates of the vertices: R(3, 3), S(-2, 3), and T(-2, -3).
To calculate the lengths of the sides, we can use the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is given by:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
Calculating the lengths of the sides RS, RT, and ST:
RS = √[(-2 - 3)^2 + (3 - 3)^2] = √[5^2 + 0^2] = √25 = 5
RT = √[(-2 - 3)^2 + (-3 - 3)^2] = √[5^2 + (-6)^2] = √[25 + 36] = √61 ≈ 7.81
ST = √[(3 - (-2))^2 + (3 - (-3))^2] = √[5^2 + 6^2] = √[25 + 36] = √61 ≈ 7.81
Now, to find the measure of the smallest angle, we can use the Law of Cosines:
cos(A) = (b² + c² - a²) / (2bc)
where A is the angle opposite side a, b is the side opposite angle B, and c is the side opposite angle C.
In this case, since we want the smallest angle, we will find the angle opposite the shortest side, RS:
cos(A) = (RT² + ST² - RS²) / (2RT * ST)
cos(A) = (7.81² + 7.81² - 5²) / (2 * 7.81 * 7.81)
cos(A) = (61 + 61 - 25) / (122.02)
cos(A) = 97 / 122.02
cos(A) ≈ 0.7943
To find the measure of angle A, we can take the inverse cosine (also known as arccosine) of the cosine value:
A ≈ arccos(0.7943)
A ≈ 38.7801º
Therefore, the smallest angle in the triangle is approximately 38.7801º. Since none of the provided options match this value, none of the given options are correct.