The following figure shows $\triangle A B C$ with side lengths to the nearest tenth.

Find $A C$ in $\triangle A B C$.
Round to the nearest tenth.
$$
A C=
$$
$\square$

Question

The following figure shows $\triangle A B C$ with side lengths to the nearest tenth.

Find $A C$ in $\triangle A B C$.
Round to the nearest tenth.
$$
A C=
$$
$\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the length of side AC, we can use the Law of Sines, which relates the lengths of sides of a triangle to the sines of its opposite angles. ‖ ‖ step 2 ⋮ The Law of Sines states that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, where $a$, $b$, and $c$ are the lengths of the sides opposite to angles $A$, $B$, and $C$ respectively. ‖ ‖ step 3 ⋮ Given that $AB = 12$ units, $\angle C = 40^\circ$, and $\angle B = 110^\circ$, we can find $\angle A$ by using the fact that the sum of angles in a triangle is $180^\circ$. ‖ ‖ step 4 ⋮ Calculating $\angle A$: $\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 110^\circ - 40^\circ = 30^\circ$. ‖ ‖ step 5 ⋮ Now we can apply the Law of Sines to find $AC$: $\frac{AC}{\sin B} = \frac{AB}{\sin A}$. ‖ ‖ step 6 ⋮ Substituting the known values into the Law of Sines: $\frac{AC}{\sin 110^\circ} = \frac{12}{\sin 30^\circ}$. ‖ ‖ step 7 ⋮ Solving for $AC$: $AC = \frac{12 \cdot \sin 110^\circ}{\sin 30^\circ}$. ‖ ‖ step 8 ⋮ Since $\sin 30^\circ = \frac{1}{2}$, the equation simplifies to: $AC = \frac{12 \cdot \sin 110^\circ}{\frac{1}{2}} = 24 \cdot \sin 110^\circ$. ‖ ‖ step 9 ⋮ Calculate $AC$ using a calculator and round to the nearest tenth: $AC \approx 24 \cdot \sin 110^\circ$. ‖ ∻Answer∻ ⚹ [Insert final answer here after calculation] ⚹ ∻Key Concept∻ ⚹Law of Sines⚹ ∻Explanation∻ ⚹The Law of Sines is used to find unknown side lengths in a triangle when given two angles and one side.⚹◊What is the formula to calculate the length of side AC in a triangle given the other side lengths and angle measures?⍭ Generate me a similar question◊

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