Question

Solve for x. Round to the nearest tenth of a degree, if necessary.
P
Q
R
17
9.8
x°
Answer
Attempt 1 out of 2

Answer 1

I can’t see the triangle from your text — the vertex/side labels got lost. Could you either upload the diagram or tell me exactly which sides and angle correspond to P, Q, R and where x is (for example “PQ = 17, PR = 9.8, angle P = x” or “side opposite P = 17, side opposite Q = 9.8, angle R = x”, etc.)?

Meanwhile, here are the two standard ways we’ll solve for x depending on the given info:

- If you know two sides and the included angle, use the Law of Cosines:
cos(C) = (a^2 + b^2 - c^2) / (2ab),
then C = arccos(...)
(round C to the nearest tenth of a degree).

- If you know two sides and a non-included angle (one angle and its opposite side, plus another side), use the Law of Sines:
sin(A)/a = sin(B)/b,
so B = arcsin( b * sin(A) / a ).
Watch for the ambiguous case (two possible angles) when using arcsin; check triangle feasibility.

Tell me which labels correspond to 17, 9.8 and x and I’ll compute x and round it to one decimal place.