A triangle has sides that measure 10 feet, 7 feet, and 12 feet. Will it be acute, obtuse, or right?
gotta show how I got my answer.
Thanks guys!
I need help please
test it with Pythagoras ... a^ + b^2 = c^2
is the sum of the squares of the two shorter sides
... more than , less than , or equal to the square of the longest side
10^2 + 7^2 = 149 ... 12^2 = 144
so the largest angle is less than 90º
To determine whether a triangle is acute, obtuse, or right, we need to look at the angles of the triangle.
To find the angles, we can use the Law of Cosines, which states that:
c^2 = a^2 + b^2 - 2ab * cos(C)
Where,
c is the length of the side across from angle C,
a and b are the lengths of the other two sides.
In this case, we have:
a = 10 feet
b = 7 feet
c = 12 feet
Using the Law of Cosines, we can find the cosine of angle C:
cos(C) = (a^2 + b^2 - c^2) / (2ab)
Substituting the values, we get:
cos(C) = (10^2 + 7^2 - 12^2) / (2 * 10 * 7)
= (100 + 49 - 144) / 140
= 5 / 14
To determine the angle, we can use the inverse cosine (arccos) function:
C = arccos(5/14)
Using a calculator, we find:
C ≈ 64.7°
Now that we know the value of angle C, we can determine the type of triangle based on its angles:
- If C is less than 90°, the triangle is acute.
- If C is equal to 90°, the triangle is right.
- If C is greater than 90°, the triangle is obtuse.
In this case, since C ≈ 64.7°, which is less than 90°, we can conclude that the triangle is acute.
Therefore, the triangle with sides measuring 10 feet, 7 feet, and 12 feet is an acute triangle.