Answer:
Congruent bases does NOT guarantee that a trapezoid is isosceles.
Given:

x lies in the III quadrant.
To find:
The values of
.
Solution:
It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.
We know that,




x lies in the III quadrant. So,


Now,



And,





We know that,



Therefore, the required values are
.
Answer:
-30, 210 and 330 degrees
Step-by-step explanation:
Given the expression Sin x = - 1/2
x = arcsin(-1/2)
x = -30 degrees
Since sin is negative in the third and fourth quadrant
x = 180 + 30
x = 210 degrees
In the fourth quadrant
x = 360 - 30
x = 330
Hence the value of x within the interval -360 < x < 360 are -30, 210 and 330 degrees
Answer:
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