Question

There are two possible triangles with the measures given for triangle ABC. b = 20.2, c = 18.3, C = 38°

Answer 1

Ok 
First use the sine rule to find one value

b / sin B = c / sin C
20.2 / sin B  =  18.3 / sin 38

sinB =  20.2 * sin 38  / 18.3   =  0.6796
m < B = 42.8 degrees

the other possible measure is 180 - 42.8 =  137.2 degrees

Answer 2

Answer:

Two possible triangle:

• $a=29.3, b=20.2, c=18.3, A=99^{\circ}, B=43^{\circ}\text{ and } C=38^{\circ}$

• $a=2.3, b=20.2, c=18.3, A=5^{\circ}, B=137^{\circ}\text{ and } C=38^{\circ}$

Step-by-step explanation:

We are given some info about triangle ABC.

b=20.2, c=18.3, C=38°

Using sine law we can find another part of triangles.

$\text{Sine law: } \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

$\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Substitute b=20.2, c=18.3, C=38° into above formula and solve for B

$\dfrac{20.2}{\sin B}=\dfrac{18.3}{\sin 38^{\circ}}$

$\sin B=\dfrac{20.2\times \sin 38^{\cic}}{18.3}$

$\sin B=0.6796$

Two possible value of B

Case 1:

$B=42.8^{\circ}\approx 43^{\circ}$

$A+B+C=180^{\circ}$            (Angle sum property of triangle)

$A=180^{\circ}-38^{\circ}-43^{\circ}=99^{\circ}$  

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$

$a=\dfrac{20.2\times \sin99^{\circ}}{\sin 43^{\circ}}\Rightarrow 29.3$

Case 2:

$B=137^{\circ}$

$A+B+C=180^{\circ}$            (Angle sum property of triangle)

$A=180^{\circ}-38^{\circ}-137^{\circ}=5^{\circ}$  

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$

$a=\dfrac{20.2\times \sin5^{\circ}}{\sin 137^{\circ}}\Rightarrow 2.3$

Two possible triangle:

• $a=29.3, b=20.2, c=18.3, A=99^{\circ}, B=43^{\circ}\text{ and } C=38^{\circ}$

• $a=2.3, b=20.2, c=18.3, A=5^{\circ}, B=137^{\circ}\text{ and } C=38^{\circ}$

Thus, Two possible triangles.