Question

math 4 - unit 4 law of sin & cos review. find each measurement indicated. round your answers to the nearest tenth.

1. find ac
2. find ac
3. find ab
4. find ac
5. find m∠b
6. find m∠b

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines for finding side \(b\) in \(\triangle ABC\) is \(b^{2}=a^{2}+c^{2}-2ac\cos B\). For the first - type of problems asking to find a side length of a triangle, we use this formula when we know two sides and the included - angle. For example, if in a triangle we know \(a\), \(c\), and \(B\), we substitute the values into the formula and then take the square - root of the result to find \(b\).

Step2: Recall the Law of Sines

The Law of Sines is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). When we know two angles and one side of a triangle and want to find another side, we use the Law of Sines. For example, if we know \(A\), \(B\), and \(a\), we can find \(b\) as \(b = \frac{a\sin B}{\sin A}\).

Step3: Recall the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\), i.e., \(A + B + C=180^{\circ}\). When we need to find an unknown angle, we can use this property if we know the other two angles.

Step4: Solve for side lengths and angles

1. For the first problem of finding \(AC\) (assuming we know two sides and the included - angle), if \(a\), \(c\), and \(B\) are known, we use \(AC^{2}=AB^{2}+BC^{2}-2(AB)(BC)\cos B\), then \(AC=\sqrt{AB^{2}+BC^{2}-2(AB)(BC)\cos B}\).
2. For finding an angle like \(m\angle B\), if we know the three sides of the triangle, we can re - arrange the Law of Cosines to \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\), then \(m\angle B=\cos^{-1}(\frac{a^{2}+c^{2}-b^{2}}{2ac})\). If we know two sides and an angle opposite one of them, we use the Law of Sines \(\frac{\sin B}{b}=\frac{\sin A}{a}\) to find \(\sin B\) and then \(B\) (taking into account the possible two - solution cases for non - right triangles).

Since the figures and values are not fully clear from the image for a definite numerical answer:

1. For finding side lengths:

• If using the Law of Cosines \(b=\sqrt{a^{2}+c^{2}-2ac\cos B}\).
• If using the Law of Sines \(b=\frac{a\sin B}{\sin A}\).

1. For finding angles:

• If using the Law of Cosines \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\), then \(B = \cos^{-1}(\frac{a^{2}+c^{2}-b^{2}}{2ac})\).
• If using the Law of Sines \(\sin B=\frac{b\sin A}{a}\), then \(B=\sin^{-1}(\frac{b\sin A}{a})\) (with appropriate consideration of cases).

Answer:

The solutions depend on the specific values of the sides and angles given in each triangle. For example, if in a triangle with sides \(a = 12\), \(c = 22\) and included - angle \(B = 58^{\circ}\), using the Law of Cosines \(b=\sqrt{12^{2}+22^{2}-2\times12\times22\times\cos58^{\circ}}\approx\sqrt{144 + 484-528\times0.5299}\approx\sqrt{144 + 484 - 279.7872}\approx\sqrt{348.2128}\approx18.7\). If using the Law of Sines, we need to know two angles and one side to find another side. And for finding angles, we use the appropriate formula as described above.