Question

law of sines (partially visible)
triangle qrs with vertex r (100°), side qr = 2.4, side qs = 3.5
law of sines: (\frac{sin(a)}{a} = \frac{sin(b)}{b} = \frac{sin(c)}{c})
question: which equation is true for triangle qrs?
options (visible):
(\frac{sin(100^circ)}{2.4} = \frac{sin(q)}{3.5})
(\frac{sin(100^circ)}{3.5} = \frac{sin(s)}{2.4})
(\frac{sin(100^circ)}{3.5} = \frac{sin(q)}{2.4})

Explanation:

Step1: Recall Law of Sines

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

Step2: Identify Angles and Sides in \(\triangle QRS\)

• Angle at \( R \) is \( 100^\circ \), opposite side is \( QS = 3.5 \).
• Angle at \( S \) (let's call it \( \angle S \)) has opposite side \( QR = 2.4 \).
• Angle at \( Q \) (let's call it \( \angle Q \)) has opposite side \( RS \) (not needed here).

Using Law of Sines: \(\frac{\sin(\angle R)}{QS}=\frac{\sin(\angle S)}{QR}\)

Substitute \( \angle R = 100^\circ \), \( QS = 3.5 \), \( \angle S = S \), \( QR = 2.4 \):

\(\frac{\sin(100^\circ)}{3.5}=\frac{\sin(S)}{2.4}\)

Answer:

\(\boldsymbol{\frac{\sin(100^\circ)}{3.5}=\frac{\sin(S)}{2.4}}\) (the second option)