Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
The solution to the question is:
c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
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
[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] -2ab cos C-----------------1
then c = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
if b is the side facing the included angle B, then
[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] -2accosB-----------------2
b = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
from equation 2, make cosB the subject of equation
2ac cosB = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] - [tex]b^{2}[/tex]
cosB = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
if a is the side facing the included angle A, then
[tex]a^{2}[/tex] = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] -2bccosA--------------------3
a = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
from equation 3, making cosA subject of the equation
2bcosA = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] - [tex]a^{2}[/tex]
cosA = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
from equation 1, making cos C the subject
2abcosC = [tex]b^{2}[/tex] + [tex]a^{2}[/tex] - [tex]c^{2}[/tex]
cos C = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
In conclusion,
c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
Learn more about cosine rule: brainly.com/question/4372174
$SPJ1
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.