As the exercise says, the triangles are similar. So, we can set up proportions between correspondent sides.
In order to solve for x we can set up the proportion between the horizontal and vertical sides:
Solving this proportion for x implies
Now you can solve for m and p using the pythagorean theorem, because both triangles are right:
Then, we know that the hypothenuse of the big triangle is m+p, so we have
which implies
Answer:
B. Arithmetric
Step-by-step explanation:
Answer:
See below.
Step-by-step explanation:
So, we have:
Recall that secant is simply the reciprocal of cosine. So we can:
Now, recall the unit circle. Since cosine is negative, it must be in Quadrants II and/or III. The numerator is the square root of 3. The denominator is 2. The whole thing is negative. Therefore, this means that 150 or 5π/6 is a candidate. Therefore, due to reference angles, 180+30=210 or 7π/6 is also a candidate.
Therefore, the possible values for theta is
5π/6 +2nπ
and
7π/6 + 2nπ