Question

Use the inverse of cosine to find the approximate interior angle measure at vertex A . Round your answer to the nearest whole degree. ( ABC triangle. AB is 17, BC is 15, CA is 8 )

Answer 1

To find the angle at vertex A in triangle ABC using the cosine rule, we can use the following formula:

\[
\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}
\]

where:
- \( a \) is the length of side BC,
- \( b \) is the length of side CA,
- \( c \) is the length of side AB.

From the problem, we have:
- \( a = 15 \) (BC),
- \( b = 8 \) (CA),
- \( c = 17 \) (AB).

Now, we will substitute the values into the formula.

First, compute \( b^2 + c^2 - a^2 \):

\[
b^2 = 8^2 = 64
\]
\[
c^2 = 17^2 = 289
\]
\[
a^2 = 15^2 = 225
\]

Now calculate \( b^2 + c^2 - a^2 \):

\[
b^2 + c^2 - a^2 = 64 + 289 - 225 = 128
\]

Next, we calculate \( 2bc \):

\[
2bc = 2 \cdot 8 \cdot 17 = 272
\]

Now we can find \( \cos(A) \):

\[
\cos(A) = \frac{128}{272}
\]
\[
\cos(A) = \frac{16}{34} \approx 0.470588
\]

Now, to find angle A, we use the inverse cosine function:

\[
A \approx \cos^{-1}(0.470588)
\]

Using a calculator to find the angle:

\[
A \approx 62.7 \text{ degrees}
\]

Rounding to the nearest whole degree, we find:

\[
\boxed{63} \text{ degrees}
\]