Mathematics: Law of sines: startfraction sine (uppercase a) over a endfraction = startfraction sine (uppercase b) over b endfraction = startfraction sine (uppercase c) over c endfraction in δabc, c = 5.4, a = 3.3, and measure of angle a = 20 degrees. what are the possible approximate lengths of b? use the law of sines to find the answer. 2.0 units and 4.6 units 2.1 units and 8.7 units 2.3 units and 7.8 units 2.6 units and 6.6 units

The answer would be 3 small notebooks and 3 large notebooks10+10+10=30 6+6+6=1830+18= 48So 48$ is the right answer...

Law of sines: startfraction sine (uppercase a)

mookiejacks

09.01.2022

First of all, this problem is properly done with the Law of Cosines, which tells us

$a^2 = b^2 + c^2 - 2 b c \cos A$

giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.

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

We have c,a,A so the Law of Sines gives us sin C

$\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597$

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:

$C_a = \arcsin(.5597) = 34.033^\circ$

$C_o = 180^\circ - C_a = 145.967^\circ$

Both of these make a valid triangle with A=20°. They give respective B's:

$B_a = 180^\circ - A - C_a = 125.967^\circ$

$B_o = 180^\circ - A - C_o = 14.033^\circ$

So we get two possibilities for b:

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

$b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8$

$b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3$

2.3 units and 7.8 units

Let's check it with the Law of Cosines:

$a^2 = b^2 + c^2 - 2 b c \cos A$

$0 = b^2 - (2 c \cos A)b + (c^2-a^2)$

There's a shortcut for the quadratic formula when the middle term is 'even.'

$b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}$

$b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}$

$b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}$

$b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark$

Looks good.

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