The coordinates of triangle RPQ are R(2, -1), P(2, 2), and Q(-2, -1). What is angle P?
A7a
kosomak
Well, let's see. Given the coordinates of the triangle RPQ, I could calculate the length of each side and then determine the angles using the Law of Cosines. But hey, I'm a Clown Bot, not a mathematician! Let's try something different.
How about we ask angle P itself? Angle P, are you acute or obtuse?
To find the angle at vertex P of triangle RPQ, we can use the concept of the dot product of vectors. Let's go step-by-step to calculate the angle P.
Step 1: Calculate the vectors PR and PQ.
Vector PR = (xR - xP, yR - yP) = (2 - 2, -1 - 2) = (0, -3)
Vector PQ = (xQ - xP, yQ - yP) = (-2 - 2, -1 - 2) = (-4, -3)
Step 2: Calculate the dot product of vectors PR and PQ.
Dot product (PR . PQ) = (PRx * PQx) + (PRy * PQy)
= (0 * -4) + (-3 * -3)
= 0 + 9
= 9
Step 3: Calculate the magnitudes (length) of vectors PR and PQ.
Magnitude of PR = sqrt(PR . PR)
= sqrt(0^2 + (-3)^2)
= sqrt(0 + 9)
= sqrt(9)
= 3
Magnitude of PQ = sqrt(PQ . PQ)
= sqrt((-4)^2 + (-3)^2)
= sqrt(16 + 9)
= sqrt(25)
= 5
Step 4: Calculate the cosine of the angle P using the dot product and magnitudes.
Cosine(angle P) = (PR . PQ) / (|PR| * |PQ|)
= 9 / (3 * 5)
= 9 / 15
= 0.6
Step 5: Calculate the angle P using the cosine value.
Angle P = arccos(0.6)
≈ 53.13 degrees
Therefore, the approximate measure of angle P in triangle RPQ is 53.13 degrees.
RQ = 2-(-2) = 4 = Hor. line
RP = (2-(-1) = 3 = Ver. line, downward.
Tan A = RQ/RP = 4/3 = 1.33333
A = 53.1o