USE THE INFORMATION TO DECIDE IF TRIANGLE ABC IS ACUTE, RIGHT, OR OBTUSE
AC = 13
BC = sq rt. 34
CD = 3
I know the answer is obtuse but I don't know how to get there. The sq rts threw me off
Your example cannot make a triangle because the longest side (13) is longer than the sum of the other two sides (5.83 + 3 = 8.31).
There is a rule that says that if you are given any two sides of a triangle, A and b, the third side must be less than A + B and greater than |A - B|
no
To determine if triangle ABC is acute, right, or obtuse, we need to examine the angles of the triangle. In particular, we can use the Law of Cosines to determine the type of triangle based on the lengths of the sides.
The Law of Cosines states that for any triangle with side lengths a, b, and c, and angle C opposite side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
In our case, we have AC = 13, BC = square root of 34 (√34), and CD = 3.
Let's denote angle C as angle BAC (the angle opposite side AC) and a = 13, b = √34, and c = 3.
We can now use the Law of Cosines to find the cosine of angle BAC:
(3)^2 = (13)^2 + (√34)^2 - 2(13)(√34) * cos(BAC)
9 = 169 + 34 - 2(13)(√34) * cos(BAC)
9 = 203 - 2(13)(√34) * cos(BAC)
2(13)(√34) * cos(BAC) = 203 - 9
2(13)(√34) * cos(BAC) = 194
Now, let's isolate cos(BAC) by dividing both sides by 2(13)(√34):
cos(BAC) = 194 / (2(13)(√34))
cos(BAC) = 194 / (26√34)
cos(BAC) ≈ 0.8284
Since we know that cosine values range between -1 and 1, and cos(BAC) is positive (approximately 0.8284), we can conclude that angle BAC is acute. Therefore, triangle ABC is an acute triangle.
Note: The square root (√) symbol was used to represent the square root of 34 in the original question.