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Whitepunk [10]
2 years ago
13

I NEED HELP PLEASE!!!

Mathematics
1 answer:
solmaris [256]2 years ago
7 0

Answer:

C

Step-by-step explanation:

cos(theta) is positive in IV quadrant and sin(theta)=sqrt(13^2-5^2)/13=-12/13

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What is one possible solution for the triangle below?
elixir [45]
Answer:

segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°

Explanation:

1) Given two side lenghts and one angle you can use sine law:

\frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c}

2) Using the sides with length 43 in and 40in, and the corresponding opposite angles, Z and 68°, that leads to:

\frac{sinZ}{43} = \frac{sin68}{40}

From which you can clear sinZ and get:

sinZ = 43 × sin(68) / 40 = 0.9967

⇒ Z = arcsine(0.9967) ≈ 85.36°


3) The third angle can be determined using 85.36° + 68° + X = 180°

⇒ X = 180° - 85.36° - 68° = 26.64°.

4) Finally, you can apply the law of sine to obtain the last missing length:

\frac{x}{sin(26.64)} = \frac{40}{sin(68)}

From which: x = 40 × sin(26.64°) /  sin(68°) = 19.34 in

The answer, then is:
segment YZ ≈ 19.4 in
angle X ≈ 85.3°
angle Z ≈ 26.7°
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