Find, correct to one decimal place, the size of the smallest angle of the triangle which has sides of length 3,5,7
a^2 = b^2 + c^2 - 2bc cosA
b^2 = a^2 + c^2 - 2ac cosB
c^2 = b^2 + a^2 - 2ba cosC
where a, b, and c are the sides of the triangle, and A, B, C are the angles opposite the sides.
Plug a, b, c = 3, 5, 7 into these equations and solve for the angles A, B, C. The smallest of these angles is the answer
you can save some work by recognizing that the smallest angle is opposite the shortest side (law of sines)
To find the size of the smallest angle of a triangle with sides 3, 5, and 7, we can use the Law of Cosines.
The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite to side c, the cosine of angle C can be found using the following formula:
cos(C) = (a^2 + b^2 - c^2) / (2ab)
In this case, we want to find the smallest angle, so we need to find the side opposite to that angle, which is 3. Let's call the other two sides b and c, where b = 5 and c = 7.
Now we can substitute the values into the Law of Cosines formula:
cos(C) = (3^2 + 5^2 - 7^2) / (2 * 3 * 5)
= (9 + 25 - 49) / (30)
= (-15) / 30
= -0.5
Note that we have obtained a negative value for the cosine of angle C. This means that angle C is greater than 90 degrees, or obtuse. However, we are interested in finding the smallest angle, which must be acute (less than 90 degrees).
To find the smallest angle, we can take the inverse cosine (or arccos) of -0.5:
smallest angle = arccos(-0.5)
Using a calculator, we can find:
smallest angle ≈ 120.0 degrees
Therefore, the size of the smallest angle in the triangle with sides 3, 5, and 7 is approximately 120.0 degrees.