Question

Could someone please help me and show work

Answer 1

Step-by-step explanation:

remember, the sum of all angles in a triangle is always 180°.

and the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

with a, b, c being the sides opposite of their associated angles (A, B, C).

1)

17/sin(26) = 12/sin(C)

sin(C) = 12×sin(26)/17 = 0.309438457...

C = 18.02539254...° ≈ 18°

3)

180 = 27 + 115 + C

C = 38°

19/sin(38) = AC/sin(27)

AC = 19×sin(27)/sin(38) = 14.01065332... in ≈ 14 in

Answer 2

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,

$\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}$, 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,

$\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}$

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,

$\frac{sin A}{a} =\frac{sin B}{b}= \frac{sinC}{c}$

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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