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In ΔIJK, j = 530 cm, i = 740 cm and ∠I=133°. Find all possible values of ∠J, to the nearest degree.

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Final answer:

Using the Law of Sines in triangle IJK, with the given side lengths and angle I, the possible values for angle J are approximately 47° and 0° when ∠I is 133°.

Step-by-step explanation:

To find all possible values of ∠J in ΔIJK given that j = 530 cm, i = 740 cm, and ∠I=133°, we can use the Law of Sines:


ü = ∑ ∠'s of a triangle = 180°

ü - ∠I = 180° - 133° = 47°

Let's call ∠K as θ. Using the Law of Sines:

√(j/sin(∠J)) = √(i/sin(θ))

sin(θ) = (i * sin(∠J))/j

sin(θ) = (740 * sin(∠J)/530

Since θ can only be as large as 47°, we use the inverse sine function to determine the maximum value of sin(∠J) that would yield sin(θ) ≤ 1. By setting sin(θ) equal to 1, we can find the maximum possible angle for ∠J. Then we can find the second possible value for ∠J by subtracting the first ∠J from 47° as both ∠J and ∠K must add up to 47°.
sin(∠J) max = 530/740

∠J max = arcsin(530/740)

∠J max ≈ 47° (to the nearest degree)

Lastly, to find the second value for ∠J, we subtract ∠J max from 47°:
∠J second = 47° - ∠J max

∠J second ≈ 47° - 47° = 0° (to the nearest degree)

Therefore, the possible values for ∠J to the nearest degree are approximately 47° and 0°.

User Muhammad Naufil
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