Asksia AI LOGO


Posted 2 months ago

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

Find ACA C in ABC\triangle A B C.
Round to the nearest tenth.
A C=

Sign in to unlock the answer
Answer from Sia
Posted 2 months ago
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 asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, where aa, bb, and cc are the lengths of the sides opposite to angles AA, BB, and CC respectively
step 3
Given that AB=12AB = 12 units, C=40\angle C = 40^\circ, and B=110\angle B = 110^\circ, we can find A\angle A by using the fact that the sum of angles in a triangle is 180180^\circ
step 4
Calculating A\angle A: A=180BC=18011040=30\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 ACAC: ACsinB=ABsinA\frac{AC}{\sin B} = \frac{AB}{\sin A}
step 6
Substituting the known values into the Law of Sines: ACsin110=12sin30\frac{AC}{\sin 110^\circ} = \frac{12}{\sin 30^\circ}
step 7
Solving for ACAC: AC=12sin110sin30AC = \frac{12 \cdot \sin 110^\circ}{\sin 30^\circ}
step 8
Since sin30=12\sin 30^\circ = \frac{1}{2}, the equation simplifies to: AC=12sin11012=24sin110AC = \frac{12 \cdot \sin 110^\circ}{\frac{1}{2}} = 24 \cdot \sin 110^\circ
step 9
Calculate ACAC using a calculator and round to the nearest tenth: AC24sin110AC \approx 24 \cdot \sin 110^\circ
[Insert final answer here after calculation]
Key Concept
Law of Sines
The Law of Sines is used to find unknown side lengths in a triangle when given two angles and one side.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages
Strong algorithms that better know you
Early access to new release features
Study Other Question