Answer:
∠CAB = 39.5°
AB = 15.3 cm
Step-by-step explanation:
In triangle ABC, we are given:
- AC = 14 cm
- BC = 10 cm
- ∠ABC = 63°
Since we have been given two sides and the angle opposite one of these given sides (SSA configuration), this is an ambiguous case leading to the possibility of two different triangles, one triangle, or no triangle at all. The possibilities are:
- If ∠ABC < 90° and AC < BC and sin CAB > 1, there are no triangles.
- If ∠ABC < 90° and AC < BC and sin CAB < 1, there are two triangles.
- If ∠ABC < 90° and AC ≥ BC, there is one triangle.
- If ∠ABC > 90° and AC ≤ BC, there are no triangles.
- If ∠ABC > 90° and AC > BC, there is one triangle.
As the given angle ABC is acute (∠ABC < 90°), and AC > BC, then one triangle is possible.
To find the measure of angle CAB, we can use the Law of Sines:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
In this case:
- A = ∠CAB
- B = ∠ABC = 63°
- C = ∠ACB
- a = BC = 10 cm
- b = AC = 14 cm
- c = AB
Substitute the given values into the formula:
[tex]\dfrac{\sin CAB}{10}=\dfrac{\sin 63^{\circ}}{14}=\dfrac{\sin ACB}{AB}[/tex]
Solve for angle CAB:
[tex]\dfrac{\sin CAB}{10}=\dfrac{\sin 63^{\circ}}{14}\\\\\\ \sin CAB=\dfrac{10\sin 63^{\circ}}{14}\\\\\\ CAB=\sin^{-1}\left( \dfrac{10\sin 63^{\circ}}{14}\right) \\\\\\ CAB=39.5263658107566...^{\circ}\\\\\\CAB=39.5^{\circ}\; \sf (nearest \; tenth)[/tex]
Therefore, the measure of angle CAB is 39.5°.
To find the measure of angle ACB, we can use the Triangle Sum Theorem, which states that the sum of the interior angles of any triangle is always 180°:
[tex]ACB+CAB+ABC=180^{\circ}\\\\ACB+39.5263658107566...^{\circ}+63^{\circ}=180^{\circ}\\\\ACB+102.5263658107566...^{\circ}=180^{\circ}\\\\ACB=180^{\circ}-102.5263658107566...^{\circ}\\\\ACB=77.47363418924...^{\circ}\\\\ACB=77.5^{\circ}\; \sf (nearest \; tenth)[/tex]
Now, use the Law of Sines again to find the length of AB, remembering to use the exact value of angle ACB:
[tex]\dfrac{\sin CAB}{10}=\dfrac{\sin 60^{\circ}}{14}=\dfrac{\sin (77.47363418924...^{\circ})}{AB} \\\\\\[/tex]
Solve for AB:
[tex]\dfrac{\sin 63^{\circ}}{14}=\dfrac{\sin (77.47363418924...^{\circ})}{AB} \\\\\\ AB \sin 63^{\circ}=14\sin (77.47363418924...^{\circ})\\\\\\AB=\dfrac{14\sin (77.47363418924...^{\circ})}{\sin 63^{\circ}}\\\\\\ AB=15.3385501617379...\\\\\\AB=15.3\; \sf cm\;(nearest\;tenth)[/tex]
Therefore, the length of AB is 15.3 cm.
