Question

use the law of sines to solve the following non - right triangles. round all answers to the nearest tenth.
1.
2.
3.
4.

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that in a triangle $\triangle ABC$ with sides $a$, $b$, $c$ and opposite - angles $A$, $B$, $C$ respectively, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Also, the sum of the interior angles of a triangle is $180^{\circ}$, i.e., $A + B + C=180^{\circ}$.

Problem 1:

Given $B = 52^{\circ}$, $C = 45^{\circ}$, and $c = 19$.

Step2: Find angle $A$

$A=180^{\circ}-(B + C)=180^{\circ}-(52^{\circ}+45^{\circ}) = 83^{\circ}$.

Step3: Use the Law of Sines to find $a$

$\frac{a}{\sin A}=\frac{c}{\sin C}$, so $a=\frac{c\sin A}{\sin C}=\frac{19\sin83^{\circ}}{\sin45^{\circ}}\approx\frac{19\times0.9925}{0.7071}\approx26.7$.

Step4: Use the Law of Sines to find $b$

$\frac{b}{\sin B}=\frac{c}{\sin C}$, so $b = \frac{c\sin B}{\sin C}=\frac{19\sin52^{\circ}}{\sin45^{\circ}}\approx\frac{19\times0.7880}{0.7071}\approx21.1$.

Problem 2:

Given $C = 110^{\circ}$, $c = 25$, and $b = 10$.

Step2: Use the Law of Sines to find $B$

$\frac{b}{\sin B}=\frac{c}{\sin C}$, so $\sin B=\frac{b\sin C}{c}=\frac{10\sin110^{\circ}}{25}=\frac{10\times0.9397}{25}=0.3759$. Then $B=\sin^{- 1}(0.3759)\approx22.1^{\circ}$.

Step3: Find angle $A$

$A = 180^{\circ}-(B + C)=180^{\circ}-(22.1^{\circ}+110^{\circ})=47.9^{\circ}$.

Step4: Use the Law of Sines to find $a$

$\frac{a}{\sin A}=\frac{c}{\sin C}$, so $a=\frac{c\sin A}{\sin C}=\frac{25\sin47.9^{\circ}}{\sin110^{\circ}}\approx\frac{25\times0.7421}{0.9397}\approx19.7$.

Problem 3:

Given $A = 33^{\circ}$, $C = 81^{\circ}$, and $c = 26$.

Step2: Find angle $B$

$B=180^{\circ}-(A + C)=180^{\circ}-(33^{\circ}+81^{\circ}) = 66^{\circ}$.

Step3: Use the Law of Sines to find $a$

$\frac{a}{\sin A}=\frac{c}{\sin C}$, so $a=\frac{c\sin A}{\sin C}=\frac{26\sin33^{\circ}}{\sin81^{\circ}}\approx\frac{26\times0.5446}{0.9877}\approx14.3$.

Step4: Use the Law of Sines to find $b$

$\frac{b}{\sin B}=\frac{c}{\sin C}$, so $b=\frac{c\sin B}{\sin C}=\frac{26\sin66^{\circ}}{\sin81^{\circ}}\approx\frac{26\times0.9135}{0.9877}\approx24.0$.

Problem 4:

Given $B = 56^{\circ}$, $b = 14$, and $c = 17$.

Step2: Use the Law of Sines to find $C$

$\frac{c}{\sin C}=\frac{b}{\sin B}$, so $\sin C=\frac{c\sin B}{b}=\frac{17\sin56^{\circ}}{14}=\frac{17\times0.8290}{14}\approx1.002$. Since the value of sine function is between - 1 and 1, there is no triangle for this set of data.

Answer:

Problem 1:

$A = 83^{\circ}$, $a\approx26.7$, $b\approx21.1$

Problem 2:

$A\approx47.9^{\circ}$, $B\approx22.1^{\circ}$, $a\approx19.7$

Problem 3:

$B = 66^{\circ}$, $a\approx14.3$, $b\approx24.0$

Problem 4:

No triangle exists.