Use the cosine function to find $A$ if $b=34 \mathrm{~m}$ and $c=41 \mathrm{~m}$ in the triangle $A B C$. \[ A=\square^{\circ} \] (Round to the nearest tenth as needed.)
Solution
Step 1 ：We are given that in triangle ABC, b = 34 m and c = 41 m. We are asked to find the angle A.
Step 2 ：We can use the cosine rule to find the angle A. The cosine rule can be written as cos(A) = (b^2 + c^2 - a^2) / 2bc. However, we don't know the length of side a.
Step 3 ：We can use the Pythagorean theorem to find the length of side a. The Pythagorean theorem states that a^2 = c^2 - b^2. Substituting the given values, we get a = \( \sqrt{41^2 - 34^2} = 22.9128784747792 \) m.
Step 4 ：We can substitute the values of a, b, and c into the cosine rule to find A. So, A = arccos[(b^2 + c^2 - a^2) / 2bc] = arccos[(34^2 + 41^2 - 22.9128784747792^2) / (2 * 34 * 41)].
Step 5 ：Calculating the above expression, we get A = 34.0 degrees.
Step 6 ：Rounding to the nearest tenth as needed, we get A = 34.0 degrees.
Step 7 ：So, the angle A in the triangle ABC is \(\boxed{34.0^{\circ}}\).
