9514 1404 393
Answer:
obtuse
Step-by-step explanation:
The law of cosines tells you ...
b² = a² +c² -2ac·cos(B)
Substituting for a²+c² using the given equation, we have ...
b² = b²·cos(B)² -2ac·cos(B)
We can subtract b² to get a quadratic in standard form for cos(B).
b²·cos(B)² -2ac·cos(B) -b² = 0
Solving this using the quadratic formula gives ...
The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...
The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.
Answer:
True
Step-by-step explanation:
Example: 1.666666...
In that case, place the bar on top of the last 6
Answer:
At first, we have 3 expressions that are equal.
This is not true.
By definition of tangent,
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
Recall the double angle identities:
sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)
cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):
sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))
and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.
<em />
We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get
sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5
Then
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)
tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)
tan(2<em>θ</em>) = 24/7
