First of all, this problem is properly done with the Law of Cosines, which tells us
[tex]a^2 = b^2 + c^2 - 2 b c \cos A[/tex]
giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.
[tex]\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}[/tex]
We have c,a,A so the Law of Sines gives us sin C
[tex]\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597[/tex]
There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:
[tex]C_a = \arcsin(.5597) = 34.033^\circ[/tex]
[tex]C_o = 180^\circ - C_a = 145.967^\circ[/tex]
Both of these make a valid triangle with A=20°. They give respective B's:
[tex]B_a = 180^\circ - A - C_a = 125.967^\circ[/tex]
[tex]B_o = 180^\circ - A - C_o = 14.033^\circ[/tex]
So we get two possibilities for b:
[tex]b = \dfrac{a \sin B}{\sin A}[/tex]
[tex]b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8[/tex]
[tex]b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3[/tex]
Answer: 2.3 units and 7.8 units
Let's check it with the Law of Cosines:
[tex]a^2 = b^2 + c^2 - 2 b c \cos A[/tex]
[tex]0 = b^2 - (2 c \cos A)b + (c^2-a^2)[/tex]
There's a shortcut for the quadratic formula when the middle term is 'even.'
[tex]b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}[/tex]
[tex]b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}[/tex]
[tex]b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}[/tex]
[tex]b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark[/tex]
Looks good.
Answer:
2.3 and 7.8 are 100% CORRECT.
Step-by-step explanation:
Just took test on Edge and got it right.
Plz mark me brainliest.