honors pre - calculus\n© 2018 kuta software llc. all rights reserved.\nlaw of sines\nsolve each triangle…

honors pre - calculus\n© 2018 kuta software llc. all rights reserved.\nlaw of sines\nsolve each triangle. round your answers to the nearest tenth.\n1) m∠b = 18°, m∠c = 10°, a = 35\n2) m∠a = 157°, m∠c = 10°, c = 23\n3) m∠a = 69°, m∠c = 99°, b = 4\n4) m∠c = 152°, b = 15, c = 34\n5) m∠b = 121°, a = 11, b = 5\n6) m∠b = 99°, m∠a = 53°, c = 10\n7) m∠a = 65°, c = 16, a = 13\n8) m∠a = 142°, c = 29, a = 15\n9) m∠a = 119°, m∠b = 50°, c = 7\n10) m∠a = 108°, c = 23, a = 25\n11) m∠b = 61°, a = 13, b = 10\n12) m∠c = 60°, b = 30, c = 29\n13)\n14)

honors pre - calculus\n© 2018 kuta software llc. all rights reserved.\nlaw of sines\nsolve each triangle. round your answers to the nearest tenth.\n1) m∠b = 18°, m∠c = 10°, a = 35\n2) m∠a = 157°, m∠c = 10°, c = 23\n3) m∠a = 69°, m∠c = 99°, b = 4\n4) m∠c = 152°, b = 15, c = 34\n5) m∠b = 121°, a = 11, b = 5\n6) m∠b = 99°, m∠a = 53°, c = 10\n7) m∠a = 65°, c = 16, a = 13\n8) m∠a = 142°, c = 29, a = 15\n9) m∠a = 119°, m∠b = 50°, c = 7\n10) m∠a = 108°, c = 23, a = 25\n11) m∠b = 61°, a = 13, b = 10\n12) m∠c = 60°, b = 30, c = 29\n13)\n14)

Answer

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$, where $a, b, c$ are the side - lengths of a triangle and $A, B, C$ are the opposite angles respectively.

Step2: Find the third angle

In a triangle, $A + B + C=180^{\circ}$. So, if two angles are known, the third angle can be found as $A = 180^{\circ}-(B + C)$, $B = 180^{\circ}-(A + C)$ or $C = 180^{\circ}-(A + B)$.

Step3: Use the Law of Sines to find side - lengths

Once an angle and its opposite side are known, and another angle is known, we can use the Law of Sines to find the other side - lengths. For example, if we know $A,a$ and $B$, then $b=\frac{a\sin B}{\sin A}$.

Let's solve the first triangle:

  1. Given $m\angle B = 18^{\circ}$, $m\angle C = 10^{\circ}$, $a = 35$.
    • First, find $m\angle A$:
      • $m\angle A=180^{\circ}-(18^{\circ}+10^{\circ}) = 152^{\circ}$.
    • Then, use the Law of Sines to find $b$ and $c$:
      • $\frac{a}{\sin A}=\frac{b}{\sin B}$, so $b=\frac{a\sin B}{\sin A}=\frac{35\sin18^{\circ}}{\sin152^{\circ}}\approx24.1$.
      • $\frac{a}{\sin A}=\frac{c}{\sin C}$, so $c=\frac{a\sin C}{\sin A}=\frac{35\sin10^{\circ}}{\sin152^{\circ}}\approx13.9$.

We can follow the same steps for the remaining triangles.

Answer:

For the first triangle: $m\angle A = 152^{\circ}$, $b\approx24.1$, $c\approx13.9$ (and similar solutions for other triangles following the above - mentioned steps).