We are asked to decide whether the following sentence is true or false.
The Law of Cosines is used to solve a triangle when two angles and any sides are known.
Let's first recall that solving a triangle means finding all unknown side lengths and angle measures of the triangle. With this in mind, we will consider a case where we know two angles and any sides of a triangle. Let's arbitrarily draw a triangle with m ∠ A= 40, m ∠ C= 80, and a= 3.
Having m ∠ A and m ∠ C, we can find the measure of ∠ B by using the Triangle Angle Sum Theorem.
40 ^(∘) + m ∠ B + 80 ^(∘) =180
⇓
m ∠ B = 60 ^(∘)
Now, we will try to solve this triangle by using the Law of Cosines.
a^2= b^2+ c^2-2 b c cos A
b^2= a^2+ c^2-2 a c cos B
c^2= a^2+ b^2-2 a b cos C
Let's substitute our known values, m ∠ A= 40, m ∠ B = 60, m ∠ C= 80, and a= 3, into these formulas and check whether they can be solved or not.
Formula
Substitution
Check
a^2= b^2+ c^2-2 b c cos A
3^2= b^2+ c^2-2 b c cos 40
*
b^2= a^2+ c^2-2 a c cos B
b^2= 3^2+ c^2-2 * 3 c cos 60
*
c^2= 3^2+ b^2-2 a b cos C
c^2= 3^2+ b^2-2 * 3 b cos 80
*
Notice that the obtained equations cannot be solved — we ended with two unknowns of one equation. Therefore, we cannot reach the solution by using the Law of Cosines here. However, we are able to use the Law of Cosines in the following cases.
If we know the measures of two sides and the included angle.
If we know the measures of the three sides.
On the other hand, we can consider the Law of Sines to see the relationships between side lengths of a triangle and the sines of the angles opposite them.
sin A/a=sin B/b=sin C/c
Let's substitute our known values from the triangle into the Law of Sines to see whether we can solve the triangle.
sin 40/3=sin 60/b=sin 80/c
Great! Let's now find the missing side lengths one at a time.
