In DeltaMNO,o=870 inches, n=550 inches and /_N=18^(@). Find all possible values of /_O, to the nearest degree. Answer:

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In  DeltaMNO,o=870 inches,  n=550 inches and  /_N=18^(@). Find all possible values of  /_O, to the nearest degree. Answer:

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Law of Sines Formula: To find the possible values of angle O, we can use the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of the opposite angle is constant for all three sides and angles in the triangle. The formula is given by:  a/sin(A) = b/sin(B) = c/sin(C)  where a, b, and c are the lengths of the sides, and A, B, and C are the opposite angles. In this case, we have side o (opposite angle O), side n (opposite angle N), and angle N. We can set up the ratio for sides o and n:  o/sin(O) = n/sin(N)  Plugging in the given values:  870/sin(O) = 550/sin(18°)  Now we need to solve for sin(O).Calculate Sine of Angle N: First, calculate the sine of angle N, which is 18 degrees. We can use a calculator for this:  sin(18°) ≈ 0.30901699...  Now we have:  870/sin(O) ≈ 550/0.30901699...  Next, we will cross-multiply to solve for sin(O).Cross-Multiply to Solve: Cross-multiplying gives us:  sin(O) ≈ (870 * 0.30901699...) / 550  Now we calculate the right side of the equation.Calculate Sin(O): Performing the calculation:  sin(O) ≈ (870 * 0.30901699) / 550 sin(O) ≈ 268.884763 / 550 sin(O) ≈ 0.4890632...  Now we have the approximate value of sin(O).Find Angle O: To find angle O, we take the inverse sine (arcsin) of the value we found for sin(O):  O ≈ arcsin(0.4890632...)  Using a calculator, we find:  O ≈ 29.21°  Since we are looking for the value to the nearest degree, we round this to:  O ≈ 29°  However, we must consider that there could be another possible value for angle O because the sine function is positive in both the first and second quadrants. To find the second possible value, we subtract our first value from 180°:  180° - 29° = 151°  So the two possible values for angle O are approximately 29° and 151°.

In $\Delta \mathrm{MNO}, o=870$ inches, $n=550$ inches and $\angle \mathrm{N}=18^{\circ}$ . Find all possible values of $\angle \mathrm{O}$ , to the nearest degree. $\newline$ Answer:

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In ΔMNO,o=870 \Delta \mathrm{MNO}, o=870 ΔMNO,o=870 inches, n=550 n=550 n=550 inches and ∠N=18∘ \angle \mathrm{N}=18^{\circ} ∠N=18∘. Find all possible values of ∠O \angle \mathrm{O} ∠O, to the nearest degree.\newlineAnswer:

In $\Delta \mathrm{MNO}, o=870$ inches, $n=550$ inches and $\angle \mathrm{N}=18^{\circ}$ . Find all possible values of $\angle \mathrm{O}$ , to the nearest degree. $\newline$ Answer: