Question

solve the triangle.
a=4.23 m c=5.54 m b=28.1°

what is the length of side b?
□ m
(simplify your answer. type an integer or a decimal. round to four decimal places if needed.)
what is the measure of angle a?
47.8°
(simplify your answer. type an integer or a decimal. round to the nearest tenth if needed.)
what is the measure of angle c?
104.1°
(simplify your answer. type an integer or a decimal. round to the nearest tenth if needed.)

Explanation:

Step1: Use Law of Cosines for side b

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(B\) opposite side \(b\), \(b^{2}=a^{2}+c^{2}-2ac\cos B\). Given \(a = 4.23\) m, \(c = 5.54\) m, and \(B=28.1^{\circ}\). First, calculate \(\cos(28.1^{\circ})\approx0.8829\). Then:
\(a^{2}=4.23^{2}=17.8929\), \(c^{2}=5.54^{2}=30.6916\), \(2ac = 2\times4.23\times5.54=46.7244\).
So \(b^{2}=17.8929 + 30.6916-46.7244\times0.8829\).
Calculate \(46.7244\times0.8829\approx41.25\). Then \(b^{2}=17.8929 + 30.6916 - 41.25=7.3345\).
Take the square root: \(b=\sqrt{7.3345}\approx2.7082\) m.

Step2: Verify angle A (optional, but to check)

Using Law of Sines: \(\frac{\sin A}{a}=\frac{\sin B}{b}\). \(\sin A=\frac{a\sin B}{b}=\frac{4.23\times\sin(28.1^{\circ})}{2.7082}\). \(\sin(28.1^{\circ})\approx0.4714\), so \(\sin A=\frac{4.23\times0.4714}{2.7082}\approx\frac{1.994}{2.7082}\approx0.736\), \(A\approx47.4^{\circ}\)? Wait, but the given answer is \(47.8^{\circ}\), maybe due to more precise calculation. However, the main task here is to find side \(b\) correctly. Wait, maybe my initial calculation of \(\cos(28.1^{\circ})\) was less precise. Let's recalculate with more precision. \(\cos(28.1^{\circ})=\cos(28^{\circ}6')\approx\cos(28.1^{\circ}) = \cos(28 + 0.1)= \cos28\cos0.1-\sin28\sin0.1\approx0.88294759 - 0.46947156\times0.00174533\approx0.88294759 - 0.000819\approx0.882128\). Then \(2ac\cos B=2\times4.23\times5.54\times0.882128\approx46.7244\times0.882128\approx41.21\). Then \(b^{2}=17.8929 + 30.6916 - 41.21=7.3745\). Then \(b=\sqrt{7.3745}\approx2.7156\)? Wait, maybe the given angle A and C are based on correct \(b\). Wait, perhaps I made a mistake. Let's use more precise calculator steps.

Using a calculator for Law of Cosines:
\(b=\sqrt{4.23^{2}+5.54^{2}-2\times4.23\times5.54\times\cos(28.1^{\circ})}\)
Calculate each term:
\(4.23^2 = 17.8929\)
\(5.54^2 = 30.6916\)
\(2\times4.23\times5.54 = 46.7244\)
\(\cos(28.1^{\circ})\approx\cos(28.1)=0.8821276\)
So \(2\times4.23\times5.54\times\cos(28.1^{\circ})=46.7244\times0.8821276\approx41.213\)
Then \(b^2=17.8929 + 30.6916 - 41.213 = 7.3715\)
\(b=\sqrt{7.3715}\approx2.7150\) m (rounded to four decimal places). Wait, maybe the initial given angle A and C are correct, but let's check with the correct \(b\).

Wait, maybe the problem is to solve the triangle, so first find \(b\) using Law of Cosines, then find angles A and C using Law of Sines or Cosines.

But the question here is to find the length of side \(b\). Let's do it precisely:

\(b = \sqrt{a^2 + c^2 - 2ac\cos B}\)

\(a = 4.23\), \(c = 5.54\), \(B = 28.1^\circ\)

\(\cos(28.1^\circ) \approx 0.88212759\)

\(a^2 = 4.23 \times 4.23 = 17.8929\)

\(c^2 = 5.54 \times 5.54 = 30.6916\)

\(2ac = 2 \times 4.23 \times 5.54 = 46.7244\)

\(2ac\cos B = 46.7244 \times 0.88212759 \approx 41.213\)

\(a^2 + c^2 - 2ac\cos B = 17.8929 + 30.6916 - 41.213 = 7.3715\)

\(b = \sqrt{7.3715} \approx 2.7150\) m (rounded to four decimal places). Wait, but maybe the system's given angle A and C are based on a more precise \(b\). Alternatively, perhaps I miscalculated. Let's use a calculator for \(\cos(28.1^\circ)\):

Using a calculator, \(\cos(28.1^\circ) \approx \cos(28.1) = 0.88212759\)

Then:

\(b^2 = 4.23^2 + 5.54^2 - 2 \times 4.23 \times 5.54 \times \cos(28.1^\circ)\)

\(= 17.8929 + 30.6916 - 2 \times 4.23 \times 5.54 \times 0.88212759\)

\(= 48.5845 - 46.7244 \times 0.88212759\)

\(= 48.5845 - 41.213\)

\(= 7.3715\)

\(b = \sqrt{7.3715} \approx 2.7150\) m. Wait, but maybe the correct answer is approximately 2.708 or 2.71. Wait, perhaps my calculation of \(\cos(28.1^\circ)\) is wrong. Let's use a calculator for \(\cos(28.1^\circ)\):

Using a scientific calculator, \(\cos(28.1^\circ) \approx 0.8821\) (correct). Then:

\(2 \times 4.23 \times 5.54 = 46.7244\)

\(46.7244 \times 0.8821 = 46.7244 \times 0.8 + 46.7244 \times 0.08 + 46.7244 \times 0.0021 = 37.3795 + 3.73795 + 0.0981 = 41.21555\)

Then \(a^2 + c^2 = 17.8929 + 30.6916 = 48.5845\)

\(48.5845 - 41.21555 = 7.36895\)

\(b = \sqrt{7.36895} \approx 2.7146\) m (rounded to four decimal places: 2.7146).

Wait, maybe the initial given angle A and C are correct, but the key is to find side \(b\) using Law of Cosines.

Answer:

The length of side \(b\) is approximately \(\boxed{2.7146}\) m (rounded to four decimal places).