Answer:
Part A) Use the identity 
Part B) 
Step-by-step explanation:
we know that
----> by trigonometric identity
we have
substitute in the expression above
solve for tan(A)
square root both sides
Remember that Angle A terminates in quadrant IV
so
The value of tan(A) is negative
<em>Find the value of cot(A)</em>
we know that
----> is the reciprocal
therefore
simplify
ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a
First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.
(ax² + bx + c = 0)/a =
x² + (b/a)x + (c/a) = 0
Isolate (c/a) by subtracting (c/a) from both sides
x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)
x² + (b/a)x = 0 - (c/a)
Add spaces
x² + (b/a)x = -c/a
Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.
x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
Simplify the left side of the equation.
x² + (b/a)x + (b/2a)² = (x + (b/2a))²
(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))
Take the square root of both sides of the equation
√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))
x + b/(2a) = (±√(b² - 4ac)/2a
Simplify. Isolate the x.
x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a
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It could not be 73º because if you add that to 120 then it will be more than 180 degrees, which is the total number of degrees in a triangle. Hope this helps!
