Question

solve for x. round your answer to the nearest tenth if necessary. figures are not necessarily drawn to scale.
(there are two triangles: one with vertices t, s, r (ts=19, sr=25, angles at t: 50°, at s: 60°) and another with vertices v, u, w (vw=45, angle at v: 81°, angle at u: 41°, side vu is x))

Explanation:

Step1: Find the third angle in triangle VWU

The sum of angles in a triangle is \(180^\circ\). So, angle at \(W\) is \(180 - 81 - 41 = 58^\circ\)? Wait, no, wait. Wait, first, let's check the first triangle to see if they are similar? Wait, no, the first triangle: angle at \(T\) is \(39^\circ\), angle at \(S\) is \(61^\circ\), so angle at \(R\) is \(180 - 39 - 61 = 80^\circ\)? Wait, no, the second triangle: angle at \(V\) is \(81^\circ\), angle at \(U\) is \(41^\circ\), so angle at \(W\) is \(180 - 81 - 41 = 58^\circ\)? Wait, maybe I made a mistake. Wait, the first triangle: \(TS = 19\), \(SR = 25\), angle at \(T\) is \(39^\circ\), angle at \(S\) is \(61^\circ\), so angle at \(R\) is \(180 - 39 - 61 = 80^\circ\). The second triangle: angle at \(V\) is \(81^\circ\), angle at \(U\) is \(41^\circ\), so angle at \(W\) is \(180 - 81 - 41 = 58^\circ\). Wait, maybe they are not similar. Wait, maybe the first triangle is a right triangle? No, \(39 + 61 = 100\), so angle at \(R\) is \(80^\circ\). Wait, maybe the problem is using the Law of Sines. Let's look at the second triangle: sides \(VW = 45\), angle at \(V\) is \(81^\circ\), angle at \(U\) is \(41^\circ\), side \(x\) is opposite angle \(W\), and side \(VW\) is opposite angle \(U\). Wait, Law of Sines: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\). So in triangle VWU, let's denote: angle \(V = 81^\circ\), angle \(U = 41^\circ\), angle \(W = 180 - 81 - 41 = 58^\circ\). Side \(VW = 45\) (opposite angle \(U = 41^\circ\)), side \(x\) (opposite angle \(W = 58^\circ\))? Wait, no, side \(x\) is \(VU\), opposite angle \(W\). Wait, \(VW\) is opposite angle \(U\), \(VU\) (x) is opposite angle \(W\), and \(WU\) is opposite angle \(V\). So by Law of Sines: \(\frac{VW}{\sin U} = \frac{VU}{\sin W}\). So \(\frac{45}{\sin 41^\circ} = \frac{x}{\sin 58^\circ}\). Then \(x = \frac{45 \times \sin 58^\circ}{\sin 41^\circ}\). Let's calculate that. \(\sin 58^\circ \approx 0.8480\), \(\sin 41^\circ \approx 0.6561\). So \(x \approx \frac{45 \times 0.8480}{0.6561} \approx \frac{38.16}{0.6561} \approx 58.2\). Wait, but maybe I messed up the angles. Wait, the first triangle: angle at \(T\) is \(39^\circ\), angle at \(S\) is \(61^\circ\), so angle at \(R\) is \(80^\circ\). The second triangle: angle at \(V\) is \(81^\circ\), angle at \(U\) is \(41^\circ\), angle at \(W\) is \(58^\circ\). Wait, maybe the first triangle is similar to the second? No, angles don't match. Wait, maybe the first triangle is a right triangle? No, \(39 + 61 = 100\), so angle at \(R\) is \(80^\circ\). Wait, maybe the problem is that the first triangle has angle \(R = 80^\circ\), and the second triangle has angle \(V = 81^\circ\), which is close, maybe a typo? Or maybe I misread the angles. Wait, the first triangle: angle at \(T\) is \(39^\circ\), angle at \(S\) is \(61^\circ\), so angle at \(R\) is \(180 - 39 - 61 = 80^\circ\). The second triangle: angle at \(V\) is \(80^\circ\) (maybe a typo, 81 instead of 80). Let's assume angle at \(V\) is \(80^\circ\). Then angle at \(W\) is \(180 - 80 - 41 = 59^\circ\). But no, the problem says 81. Wait, maybe the first triangle is to find angle at \(R\) as \(80^\circ\), and the second triangle has angle at \(V\) as \(80^\circ\) (maybe a typo). Alternatively, maybe the two triangles are similar. Let's check the first triangle: angles \(39^\circ\), \(61^\circ\), \(80^\circ\). Second triangle: angles \(81^\circ\), \(41^\circ\), \(58^\circ\). Not similar. Wait, maybe the first triangle is a right triangle? No, \(39 + 61 = 100\), so angle at \(R\) is \(80^\circ\). Wait, maybe the problem is that the first triangle is given to find angle at \(R\) as \(80^\circ\), and then the second triangle has angle at \(V\) as \(80^\circ\) (maybe a typo in the problem, 81 instead of 80). Let's proceed with the Law of Sines as per the second triangle. So:

Step2: Apply Law of Sines to triangle VWU

Law of Sines: \(\frac{VW}{\sin U} = \frac{VU}{\sin W}\)

Given: \(VW = 45\), \(\angle U = 41^\circ\), \(\angle V = 81^\circ\), so \(\angle W = 180 - 81 - 41 = 58^\circ\)

So, \(\frac{45}{\sin 41^\circ} = \frac{x}{\sin 58^\circ}\)

Solve for \(x\):

\(x = \frac{45 \times \sin 58^\circ}{\sin 41^\circ}\)

Calculate \(\sin 58^\circ \approx 0.8480\), \(\sin 41^\circ \approx 0.6561\)

\(x \approx \frac{45 \times 0.8480}{0.6561} \approx \frac{38.16}{0.6561} \approx 58.2\) (rounded to nearest tenth)

Wait, but maybe I made a mistake in identifying the sides and angles. Let's recheck. In triangle VWU, vertices are V, W, U. So side opposite V is WU, opposite W is VU (x), opposite U is VW (45). Angles: at V is 81°, at U is 41°, so at W is 58°. So Law of Sines: \(\frac{VW}{\sin U} = \frac{VU}{\sin W}\) is correct. So the calculation seems right.

Answer:

\(x \approx 58.2\)