Question

Match each value with its formula for ABC.

Answer 1

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

What is cosine rule?

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c =  $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b =  $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB =  $a^{2}$ +  $c^{2}$ - $b^{2}$

cosB =  $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a =  $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA =  $b^{2}$ +  $c^{2}$  - $a^{2}$

cosA =  $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC =  $b^{2}$ + $a^{2}$ -  $c^{2}$

cos C =  $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

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