Question

remember, even if you enter an answer rounded to a set number of decimal places, if you use that number in a future calculation, you should use all of the decimal places reported on your calculator!
solve triangle ( abc ) if ( angle a = 41.3^{circ} ), ( a = 181.7 ), and ( b = 248.6 ).
( sin b=)
(round answer to 5 decimal places)
there are two possible angles ( b ) between ( 0^{circ} ) and ( 180^{circ} ) with this value for sine. find the two angles, and report them so that ( angle b_1 ) is the acute angle.
( angle b_1=) and ( angle b_2=)
(round these and all remaining answers to 1 decimal place)
thus, two triangles satisfy the given conditions: triangle ( a_1b_1c_1 ) and triangle ( a_2b_2c_2 ).
solve the first triangle: ( a_1b_1c_1 )
( angle c_1=) and ( c_1=)
solve the second triangle: ( a_2b_2c_2 )
( angle c_2=) and ( c_2=)

Explanation:

Step1: Use the Law of Sines to find $\sin B$

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}$.
We are given $A = 41.3^{\circ}$, $a = 181.7$, and $b = 248.6$.
Substituting these values into the Law of Sines formula:
$\sin B=\frac{b\sin A}{a}$
$\sin B=\frac{248.6\times\sin(41.3^{\circ})}{181.7}$
Using a calculator, $\sin(41.3^{\circ})\approx0.66033$
$\sin B=\frac{248.6\times0.66033}{181.7}\approx0.90909$

Step2: Find the two possible angles $B$

Since $\sin B = 0.90909$, we know that $B=\sin^{- 1}(0.90909)$ or $B = 180^{\circ}-\sin^{-1}(0.90909)$
Using a calculator, $\sin^{-1}(0.90909)\approx65.3^{\circ}$
So, $B_1 = 65.3^{\circ}$ and $B_2=180^{\circ}-65.3^{\circ}=114.7^{\circ}$

Step3: Find $C_1$ and $c_1$ for the first triangle

For triangle $A_1B_1C_1$, we know that the sum of angles in a triangle is $180^{\circ}$. So, $C_1=180^{\circ}-A - B_1$
$C_1=180^{\circ}-41.3^{\circ}-65.3^{\circ}=73.4^{\circ}$
Using the Law of Sines again, $\frac{a}{\sin A}=\frac{c_1}{\sin C_1}$
$c_1=\frac{a\sin C_1}{\sin A}$
Substituting $a = 181.7$, $A = 41.3^{\circ}$, and $C_1 = 73.4^{\circ}$
$\sin(73.4^{\circ})\approx0.95836$
$c_1=\frac{181.7\times0.95836}{0.66033}\approx263.9$

Step4: Find $C_2$ and $c_2$ for the second triangle

For triangle $A_2B_2C_2$, $C_2=180^{\circ}-A - B_2$
$C_2=180^{\circ}-41.3^{\circ}-114.7^{\circ}=24.0^{\circ}$
Using the Law of Sines, $\frac{a}{\sin A}=\frac{c_2}{\sin C_2}$
$\sin(24.0^{\circ})\approx0.40674$
$c_2=\frac{181.7\times0.40674}{0.66033}\approx111.9$

Answer:

$\sin B\approx0.90909$; $B_1 = 65.3^{\circ}$, $B_2 = 114.7^{\circ}$; $C_1 = 73.4^{\circ}$, $c_1\approx263.9$; $C_2 = 24.0^{\circ}$, $c_2\approx111.9$