Question

law of cosines
△def is an isosceles triangle with base angles e and f.
what is the angle measure of the smallest angle in the triangle?
approximate to the nearest degree.
20 × degrees
what are the measures of the two congruent base angles?
80 × degrees

Explanation:

Step1: Identify the triangle sides

In isosceles \(\triangle DEF\), \(DE = DF = 12\) (the equal sides), and \(EF = 3\) (the base). Let's denote the vertex angle (at \(D\)) as \(\alpha\) and the base angles (at \(E\) and \(F\)) as \(\beta\) (since \(E\) and \(F\) are congruent).

Step2: Apply the Law of Cosines to find the vertex angle \(\alpha\)

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and opposite angle \(C\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). For \(\triangle DEF\), to find \(\alpha\) (opposite side \(EF = 3\)), we have:
\[
EF^{2}=DE^{2}+DF^{2}-2\cdot DE\cdot DF\cdot\cos(\alpha)
\]
Substitute \(EF = 3\), \(DE = 12\), \(DF = 12\):
\[
3^{2}=12^{2}+12^{2}-2\cdot12\cdot12\cdot\cos(\alpha)
\]
\[
9 = 144 + 144-288\cos(\alpha)
\]
\[
9=288 - 288\cos(\alpha)
\]
Subtract 288 from both sides:
\[
9 - 288=- 288\cos(\alpha)
\]
\[
-279=-288\cos(\alpha)
\]
Divide both sides by \(- 288\):
\[
\cos(\alpha)=\frac{279}{288}\approx0.96875
\]
Then \(\alpha=\arccos(0.96875)\approx14^{\circ}\) (using a calculator to find the arccosine).

Step3: Find the base angles \(\beta\)

We know that the sum of angles in a triangle is \(180^{\circ}\), so \(\alpha + 2\beta=180^{\circ}\). Substitute \(\alpha\approx14^{\circ}\):
\[
14^{\circ}+2\beta = 180^{\circ}
\]
\[
2\beta=180^{\circ}- 14^{\circ}=166^{\circ}
\]
\[
\beta=\frac{166^{\circ}}{2} = 83^{\circ}
\]

Now, comparing the angles: \(\alpha\approx14^{\circ}\) and \(\beta = 83^{\circ}\), so the smallest angle is approximately \(14^{\circ}\), and the base angles are approximately \(83^{\circ}\) each.

Answer:

• The angle measure of the smallest angle: \(\boldsymbol{14}\) degrees.
• The measures of the two congruent base angles: \(\boldsymbol{83}\) degrees (each).