The basic dimensions of any triangle are its area, base, and height.
They're related by the area equation; the area is half of the base times the height, or:
We know the area and the height, so the missing dimension has to be the base. Let's set up an equation by substituting the values the problem gives into the area equation.
Can you solve this equation for
?
Answer:
120
Step-by-step explanation:
Answer:
Step-by-step explanation:
a. sin^-1(sin theta) = theta
(1/sin)(sin theta) = theta
<em>the sines cross out</em> theta = theta
b. cos(cos^-1x) = x
cos(x/cos) = x
<em>the cosines cross out </em>x = x
Answer: See Below
<u>Step-by-step explanation:</u>
NOTE: You need the Unit Circle to answer these (attached)
5) cos (t) = 1
Where on the Unit Circle does cos = 1?
Answer: at 0π (0°) and all rotations of 2π (360°)
In radians: t = 0π + 2πn
In degrees: t = 0° + 360n
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Where on the Unit Circle does
<em>Hint: sin is only positive in Quadrants I and II</em>
In degrees: t = 30° + 360n and 150° + 360n
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Where on the Unit Circle does
<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>
In degrees: t = 120° + 360n and 300° + 360n