Question

applying the law of sines
which expression gives the exact value of t?
triangle with vertex t (angle 34°), bottom - left angle 118°, right - side length 21.3; options include \\(\frac{\sin(34^{\circ})}{21.3\sin(118^{\circ})}\\) and \\(\frac{21.3\sin(34^{\circ})}{\sin(118^{\circ})}\\)

Explanation:

Step1: Recall Law of Sines

The Law of Sines states that for a triangle with sides \(a\), \(b\), \(c\) opposite angles \(A\), \(B\), \(C\) respectively, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).

Step2: Identify Angles and Sides

In the given triangle, let's label the triangle: let the angle at the bottom left be \(118^\circ\), the angle at the top \(T\) be \(34^\circ\), and the side opposite the \(118^\circ\) angle is \(21.3\), and we want to find the side \(t\) (let's assume \(t\) is opposite the \(34^\circ\) angle? Wait, no, wait. Wait, let's check the triangle. Wait, the side labeled \(21.3\) is opposite the angle of \(118^\circ\)? Wait, no, let's re - label. Let's denote: angle \(A = 118^\circ\), side \(a=t\)? Wait, no, maybe I got it wrong. Wait, let's see: the side with length \(21.3\) is adjacent to angle \(T = 34^\circ\) and angle \(118^\circ\). Let's find the third angle first. The sum of angles in a triangle is \(180^\circ\). So the third angle (let's say angle \(B\)) is \(180-(118 + 34)=28^\circ\)? Wait, no, maybe I mislabeled. Wait, actually, in the Law of Sines, if we have a triangle with angles \(A\), \(B\), \(C\) and opposite sides \(a\), \(b\), \(c\). Let's assume: angle at the bottom left is \(118^\circ\) (let's call this angle \(C = 118^\circ\)), angle at the top \(T\) is \(A=34^\circ\), then the third angle \(B=180-(118 + 34)=28^\circ\). Wait, but the side given is \(21.3\), let's say the side opposite angle \(B\) is \(21.3\)? No, maybe the side of length \(21.3\) is opposite angle \(A = 34^\circ\)? No, that can't be. Wait, maybe the correct labeling is: let the side with length \(21.3\) be opposite the angle of \(118^\circ\)? No, let's start over.

Wait, the Law of Sines formula is \(\frac{\text{side}}{\sin(\text{opposite angle})}\). Let's assume that we have a triangle where one side is \(21.3\), and the angle opposite to it is, let's see, the angle at the top is \(34^\circ\)? No, the angle at the bottom left is \(118^\circ\), the side of length \(21.3\) is on the right side. Let's denote: let the side we want to find be \(t\) (the left side), the side given is \(21.3\) (right side), the angle opposite to \(21.3\) is \(118^\circ\)? No, the angle opposite to the right side (length \(21.3\)) is the angle at the bottom left, which is \(118^\circ\)? Wait, no, the angle at the bottom left is \(118^\circ\), the side opposite to it would be the side opposite, which is the side from \(T\) to the bottom right. Wait, maybe I made a mistake. Let's use the Law of Sines correctly.

Let’s denote:

• Let angle \(T=34^\circ\), angle at the bottom left \(=118^\circ\), so the third angle (at the bottom right) \(=180 - 34-118 = 28^\circ\).
• Let the side opposite angle \(T\) ( \(34^\circ\)) be \(x\), the side opposite the bottom left angle (\(118^\circ\)) be \(y\), and the side opposite the bottom right angle (\(28^\circ\)) be \(21.3\). Wait, no, the side labeled \(21.3\) is on the right side, so maybe the side of length \(21.3\) is opposite the angle of \(34^\circ\)? No, that doesn't make sense. Wait, maybe the correct approach is: Let's say we have triangle \(ABC\), where \(A = 34^\circ\), \(B=118^\circ\), then \(C=28^\circ\), and the side opposite \(B\) ( \(118^\circ\)) is \(21.3\), and we want to find the side opposite \(A\) ( \(34^\circ\)), which is \(t\).

By Law of Sines: \(\frac{t}{\sin A}=\frac{21.3}{\sin B}\)

So \(t=\frac{21.3\sin A}{\sin B}\)

Since \(A = 34^\circ\) and \(B = 118^\circ\), then \(t=\frac{21.3\sin(34^\circ)}{\sin(118^\circ)}\)

Answer:

\(\frac{21.3\sin(34^{\circ})}{\sin(118^{\circ})}\) (the second option)