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In ΔNOP, o = 4.6 inches, n = 5.2 inches and ∠N=21°. Find all possible values of ∠O, to the nearest 10th of a degree.

Sagot :

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Here, we use the sine rule.

The sine rule states that the ratio of the sides to the sine of the angles facing that side is a constant. So, in ang given triangle with angles A, B and C and sides a, b and c respectively,

a/sinA = b/sinB = c/sinC

Using the sine rule in ΔNOP, with ∠N opposite to side n and ∠O opposite to side o.

So, o/sin∠O = n/sin∠N

Making ∠O subject of the formula, we have

∠O = sin⁻¹(osin∠N/n)

Since o = 4.6 inches, n = 5.2 inches and ∠N = 21°, substituting the values of the variable into the equation, we have

∠O = sin⁻¹(osin∠N/n)

∠O = sin⁻¹(4.6 × sin21°/5.2)

∠O = sin⁻¹(4.6 × 0.3584/5.2)

∠O = sin⁻¹(1.6485/5.2)

∠O = sin⁻¹(0.3170)

∠O = 18.5°

The other values of ∠O are the values in the second quadrant as sin∠O is positive in the first and second quadrant.

So, in the second quadrant ∠O = 180° - 18.5° = 161.5°

So, all possible values of ∠O, to the nearest 10th of a degree are 18.5° and 161.5°

Learn more about the sine rule here:

https://brainly.com/question/10080256

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