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bixtya [17]
4 years ago
13

Help me please thanks

Mathematics
1 answer:
otez555 [7]4 years ago
3 0

Answer:

B I THING I MIGHT GET IT WRONG!!!

Step-by-step explanation:


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A cyclist traveled 15 kilometers per hour faster than an in-line skater. In the time it took the cyclist to travel 12 kilometers
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Boston can be very cold at -3 degrees Fahrenheit. If the temperature continues to drop at a constant rate for four days, what is
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Which of the following is the equation of the line that passes through the point (-6, 1) and has
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Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF
Vsevolod [243]

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

Answer: 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.

6 0
3 years ago
Read 2 more answers
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