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In ΔIJK, i = 3.3 inches, k = 7 inches and ∠K=117°. Find all possible values of ∠I, to the nearest 10th of a degree.

Question

In ΔIJK, i = 3.3 inches, k = 7 inches and ∠K=117°. Find all possible values of ∠I, to the nearest 10th of a degree.

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Solution 1

To solve this problem, we will use the Law of Cosines. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and an angle γ opposite side c, the following relationship holds:

c² = a² + b² - 2ab cos(γ)

In this case, we know the lengths of sides i and k, and the measure of ∠K. We can rearrange the Law of Cosines to solve for cos(γ):

cos(γ) = (a² + b² - c²) / (2ab)

Substituting the given values:

cos(∠I) = (i² + k² - j²) / (2ik) cos(∠I) = (3.3² + 7² - j²) / (2 * 3.3 * 7)

We don't know the length of side j, but we do know that ∠K = 117°. We can use the Law of Sines to find j:

sin(∠K) = j / k j = k * sin(∠K) j = 7 * sin(117°)

Substitute j back into the equation for cos(∠I):

cos(∠I) = (3.3² + 7² - (7 * sin(117°))²) / (2 * 3.3 * 7)

Solve for ∠I:

∠I = arccos[((3.3² + 7² - (7 * sin(117°))²) / (2 * 3.3 * 7))]

This will give you the measure of ∠I in degrees. Note that because the arccos function returns values between 0° and 180°, there may be two possible values for ∠I. To find the second possible value, subtract the first value from 180°.

This problem has been solved

Solution 2

To solve this problem, we will use the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

  1. First, we find the sine of ∠K: sin(117°) = 0.8910 (rounded to the nearest four decimal places).

  2. Next, we set up the ratio of side k to the sine of ∠K, and set it equal to the ratio of side i to the sine of ∠I:

    7/sin(117°) = 3.3/sin(∠I)

  3. Solving for sin(∠I), we get:

    sin(∠I) = 3.3 * sin(117°) / 7 = 0.4180 (rounded to the nearest four decimal places).

  4. Now, we find ∠I by taking the inverse sine (also known as arcsin) of 0.4180:

    ∠I = arcsin(0.4180) = 24.9° (rounded to the nearest tenth of a degree).

  5. However, there is another possible value for ∠I. Since the sine function has a period of 360° and is positive in both the first and second quadrants, ∠I could also be 180° - 24.9° = 155.1°.

So, the possible values for ∠I are 24.9° and 155.1°.

This problem has been solved

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