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Popular Trigonometry >

solvefor b,a=arccos((a^2-b^2-c^2)/(-2bc))

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Solution

solvefor

Solution

b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​
Solution steps
a=arccos(−2bca2−b2−c2​)
Switch sidesarccos(−2bca2−b2−c2​)=a
Apply trig inverse properties
arccos(−2bca2−b2−c2​)=a
arccos(x)=a⇒x=cos(a)−2bca2−b2−c2​=cos(a)
−2bca2−b2−c2​=cos(a)
Solve −2bca2−b2−c2​=cos(a):b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​;c=0
−2bca2−b2−c2​=cos(a)
Simplify −2bca2−b2−c2​:−2bca2−b2−c2​
−2bca2−b2−c2​
Apply the fraction rule: −ba​=−ba​=−2bca2−b2−c2​
−2bca2−b2−c2​=cos(a)
Multiply both sides by b
−2bca2−b2−c2​=cos(a)
Multiply both sides by b−2bca2−b2−c2​b=cos(a)b
Simplify−2ca2−b2−c2​=cos(a)b
−2ca2−b2−c2​=cos(a)b
Solve −2ca2−b2−c2​=cos(a)b:b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​
−2ca2−b2−c2​=cos(a)b
Multiply both sides by 2c
−2ca2−b2−c2​=cos(a)b
Multiply both sides by 2c−2ca2−b2−c2​⋅2c=cos(a)b⋅2c;c=0
Simplify−(a2−b2−c2)=2bccos(a);c=0
−(a2−b2−c2)=2bccos(a);c=0
Expand −(a2−b2−c2):−a2+b2+c2
−(a2−b2−c2)
Distribute parentheses=−a2−(−b2)−(−c2)
Apply minus-plus rules−(−a)=a,−(a)=−a=−a2+b2+c2
−a2+b2+c2=2bccos(a);c=0
Move 2bccos(a)to the left side
−a2+b2+c2=2bccos(a);c=0
Subtract 2bccos(a) from both sides−a2+b2+c2−2bccos(a)=2bccos(a)−2bccos(a);c=0
Simplify−a2+b2+c2−2bccos(a)=0;c=0
−a2+b2+c2−2bccos(a)=0;c=0
Write in the standard form ax2+bx+c=0b2−2ccos(a)b−a2+c2=0;c=0
Solve with the quadratic formula
b2−2ccos(a)b−a2+c2=0
Quadratic Equation Formula:
For a=1,b=−2ccos(a),c=−a2+c2b1,2​=2⋅1−(−2ccos(a))±(−2ccos(a))2−4⋅1⋅(−a2+c2)​​
b1,2​=2⋅1−(−2ccos(a))±(−2ccos(a))2−4⋅1⋅(−a2+c2)​​
Simplify (−2ccos(a))2−4⋅1⋅(−a2+c2)​:2c2cos2(a)+a2−c2​
(−2ccos(a))2−4⋅1⋅(−a2+c2)​
(−2ccos(a))2=22c2cos2(a)
(−2ccos(a))2
Apply exponent rule: (−a)n=an,if n is even(−2ccos(a))2=(2ccos(a))2=(2ccos(a))2
Apply exponent rule: (a⋅b)n=anbn=22c2cos2(a)
4⋅1⋅(−a2+c2)=4(−a2+c2)
4⋅1⋅(−a2+c2)
Multiply the numbers: 4⋅1=4=4(c2−a2)
=22c2cos2(a)−4(c2−a2)​
Factor 22c2cos2(a)−4(−a2+c2):4(c2cos2(a)+a2−c2)
22c2cos2(a)−4(−a2+c2)
Rewrite as=4c2cos2(a)−4(−a2+c2)
Factor out common term 4=4(c2cos2(a)−(−a2+c2))
Expand c2cos2(a)−(c2−a2):c2cos2(a)+a2−c2
c2cos2(a)−(−a2+c2)
−(−a2+c2):a2−c2
−(−a2+c2)
Distribute parentheses=−(−a2)−c2
Apply minus-plus rules−(−a)=a,−(a)=−a=a2−c2
=c2cos2(a)+a2−c2
=4(c2cos2(a)−c2+a2)
=4(c2cos2(a)+a2−c2)​
Apply radical rule: assuming a≥0,b≥0=4​c2cos2(a)−c2+a2​
4​=2
4​
Factor the number: 4=22=22​
Apply radical rule: 22​=2=2
=2c2cos2(a)−c2+a2​
b1,2​=2⋅1−(−2ccos(a))±2c2cos2(a)+a2−c2​​
Separate the solutionsb1​=2⋅1−(−2ccos(a))+2c2cos2(a)+a2−c2​​,b2​=2⋅1−(−2ccos(a))−2c2cos2(a)+a2−c2​​
b=2⋅1−(−2ccos(a))+2c2cos2(a)+a2−c2​​:ccos(a)+c2cos2(a)−c2+a2​
2⋅1−(−2ccos(a))+2c2cos2(a)+a2−c2​​
Apply rule −(−a)=a=2⋅12ccos(a)+2c2cos2(a)+a2−c2​​
Multiply the numbers: 2⋅1=2=22ccos(a)+2c2cos2(a)−c2+a2​​
Factor out common term 2=22(ccos(a)+a2−c2+cos2(a)c2​)​
Divide the numbers: 22​=1=ccos(a)+c2cos2(a)−c2+a2​
b=2⋅1−(−2ccos(a))−2c2cos2(a)+a2−c2​​:ccos(a)−c2cos2(a)−c2+a2​
2⋅1−(−2ccos(a))−2c2cos2(a)+a2−c2​​
Apply rule −(−a)=a=2⋅12ccos(a)−2c2cos2(a)+a2−c2​​
Multiply the numbers: 2⋅1=2=22ccos(a)−2c2cos2(a)−c2+a2​​
Factor out common term 2=22(ccos(a)−a2−c2+cos2(a)c2​)​
Divide the numbers: 22​=1=ccos(a)−c2cos2(a)−c2+a2​
The solutions to the quadratic equation are:b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​
b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​;c=0
b=ccos(a)+c2cos2(a)−c2+a2​,b=ccos(a)−c2cos2(a)−c2+a2​

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