Question

find the length of side c.
a
b 110° 30° c
12
a = 40° c = ? b =

Explanation:

Step1: Identify Law of Sines

In a triangle, the Law of Sines states $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Here, side opposite angle \(A\) is \(a = 12\) (wait, no: side \(BC = 12\) is opposite angle \(A\), so \(a = 12\), angle \(A = 40^\circ\), angle \(C = 30^\circ\), side \(c\) is opposite angle \(C\)? Wait, no: side \(c\) is \(AB\), opposite angle \(C\) (angle at \(C\) is \(30^\circ\)), side \(a = BC = 12\) opposite angle \(A = 40^\circ\). Wait, let's clarify:

In triangle \(ABC\), vertices: \(A\), \(B\), \(C\). Angles: \(A = 40^\circ\), \(B = 110^\circ\), \(C = 30^\circ\). Sides: \(a\) is opposite \(A\) (so \(a = BC = 12\)), \(b\) is opposite \(B\) ( \(AC = b\) ), \(c\) is opposite \(C\) ( \(AB = c\) ).

So Law of Sines: $\frac{a}{\sin A}=\frac{c}{\sin C}$

Step2: Substitute values

We know \(a = 12\), \(A = 40^\circ\), \(C = 30^\circ\). So:

$\frac{12}{\sin 40^\circ}=\frac{c}{\sin 30^\circ}$

Solve for \(c\):

\(c=\frac{12 \times \sin 30^\circ}{\sin 40^\circ}\)

We know \(\sin 30^\circ = 0.5\), \(\sin 40^\circ \approx 0.6428\)

Step3: Calculate numerator and denominator

Numerator: \(12 \times 0.5 = 6\)

Denominator: \(\sin 40^\circ \approx 0.6428\)

So \(c=\frac{6}{0.6428}\approx 9.33\) (rounded to two decimal places, or as per instruction "rounded to the nearest tenth" maybe? Wait, let's check calculation again.

Wait, \(\sin 40^\circ \approx 0.6427876097\), so \(6 / 0.6427876097 \approx 9.33\), which is approximately 9.3 when rounded to the nearest tenth, or 9.33 as is.

Answer:

\(c \approx 9.3\) (or 9.33, depending on rounding)