Question

consider a triangle \\(abc\\) like the one below. suppose that \\(a = 46\\), \\(b = 62\\), and \\(c = 45\\). (the figure is not drawn to scale.) solve the triangle.

carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

if there is more than one solution, use the button labeled \or\.

\\(a = \box^\circ\\), \\(b = \box^\circ\\), \\(c = \box^\circ\\)

Explanation:

State the given values and formulas

We are given the three side lengths of triangle \(ABC\):

• \(a = 46\)
• \(b = 62\)
• \(c = 45\)

We need to find the angles \(A\), \(B\), and \(C\) using the Law of Cosines and Solving Triangles.

Calculate angle A

Using the Law of Cosines:

$$ a^2 = b^2 + c^2 - 2bc \cos A $$

$$ \cos A = \frac{b^2 + c^2 - a^2}{2bc} $$

Substitute the given values:

$$ \cos A = \frac{62^2 + 45^2 - 46^2}{2 \cdot 62 \cdot 45} = \frac{3844 + 2025 - 2116}{5580} = \frac{3753}{5580} \approx 0.6725806 $$

$$ A = \arccos(0.6725806) \approx 47.7349^\circ \approx 47.7^\circ $$

Calculate angle B

Using the Law of Cosines:

$$ b^2 = a^2 + c^2 - 2ac \cos B $$

$$ \cos B = \frac{a^2 + c^2 - b^2}{2ac} $$

Substitute the given values:

$$ \cos B = \frac{46^2 + 45^2 - 62^2}{2 \cdot 46 \cdot 45} = \frac{2116 + 2025 - 3844}{4140} = \frac{297}{4140} \approx 0.0717391 $$

$$ B = \arccos(0.0717391) \approx 85.8847^\circ \approx 85.9^\circ $$

Calculate angle C

Using the angle sum property of a triangle:

$$ C = 180^\circ - A - B $$

Substitute the intermediate values:

$$ C \approx 180^\circ - 47.7349^\circ - 85.8847^\circ = 46.3804^\circ \approx 46.4^\circ $$

Answer:

Consider a triangle \(ABC\) like the one below. Suppose that \(a = 46\), \(b = 62\), and \(c = 45\). Solve the triangle.

\(A =\) <blank>\(47.7\)</blank>\(^\circ\), \(B =\) <blank>\(85.9\)</blank>\(^\circ\), \(C =\) <blank>\(46.4\)</blank>\(^\circ\)