Respuesta :
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
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:
Two possible triangle:
- [tex]a=29.3, b=20.2, c=18.3, A=99^{\circ}, B=43^{\circ}\text{ and } C=38^{\circ}[/tex]
- [tex]a=2.3, b=20.2, c=18.3, A=5^{\circ}, B=137^{\circ}\text{ and } C=38^{\circ}[/tex]
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.
[tex]\text{Sine law: } \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
Substitute b=20.2, c=18.3, C=38° into above formula and solve for B
[tex]\dfrac{20.2}{\sin B}=\dfrac{18.3}{\sin 38^{\circ}}[/tex]
[tex]\sin B=\dfrac{20.2\times \sin 38^{\cic}}{18.3}[/tex]
[tex]\sin B=0.6796[/tex]
Two possible value of B
Case 1:
[tex]B=42.8^{\circ}\approx 43^{\circ}[/tex]
[tex]A+B+C=180^{\circ}[/tex] (Angle sum property of triangle)
[tex]A=180^{\circ}-38^{\circ}-43^{\circ}=99^{\circ}[/tex]
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}[/tex]
[tex]a=\dfrac{20.2\times \sin99^{\circ}}{\sin 43^{\circ}}\Rightarrow 29.3[/tex]
Case 2:
[tex]B=137^{\circ}[/tex]
[tex]A+B+C=180^{\circ}[/tex] (Angle sum property of triangle)
[tex]A=180^{\circ}-38^{\circ}-137^{\circ}=5^{\circ}[/tex]
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}[/tex]
[tex]a=\dfrac{20.2\times \sin5^{\circ}}{\sin 137^{\circ}}\Rightarrow 2.3[/tex]
Two possible triangle:
- [tex]a=29.3, b=20.2, c=18.3, A=99^{\circ}, B=43^{\circ}\text{ and } C=38^{\circ}[/tex]
- [tex]a=2.3, b=20.2, c=18.3, A=5^{\circ}, B=137^{\circ}\text{ and } C=38^{\circ}[/tex]
Thus, Two possible triangles.
