In first triangle side angle m∠C is 18° and in first triangle side AC is 14 in. This can be obtained by using the law of sine.
Find the required angle and side:
We know the law of sine which is,
If in a triangle, Δ ABC,
[tex]\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}[/tex], where a is the side opposite to the angle m∠A, b is the side opposite to the angle m∠B and c is the side opposite to the angle m∠C.
sine of the angle scan be found using calculator.
From the question,
1) In first triangle side,
AB = 12ft, BC = 17 ft and m∠A=26°
⇒By using the law of sine,
[tex]\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}[/tex]
sin 26°/17 = sin B/AC = sin C/12
sin 26°/17 = sin C/12
12×sin 26° = 17×sin C
sin C = 5.26/17 = 0.3091
C = 18.023° ≈ 18°
3) In second triangle side,
m∠B = 27°, AB = 19 in and m∠A = 115°
m∠A + m∠B +m∠C = 180°
115° + 27° + m∠C = 180°
142° + m∠C = 180°
m∠C = 38°
⇒By using the law of sine,
[tex]\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}[/tex]
sin 115°/BC = sin 27°/AC = sin 38°/19
sin 27°/AC = sin 38°/19
sin 27° × 19 = sin 38° × AC
8.63 = 0.62 × AC
AC = 13.91 in ≈ 14 in.
Hence in first triangle side angle m∠C is 18° and in first triangle side AC is 14 in.
Learn more about law of sine here:
brainly.com/question/17289163
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