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In /_\OPQ,p=39 inches, o=66 inches and /_O=81^(@). Find all possible values of /_P, to the nearest degree. Answer:

In △OPQ,p=39 \triangle \mathrm{OPQ}, p=39 inches, o=66 o=66 inches and ∠O=81∘ \angle \mathrm{O}=81^{\circ} . Find all possible values of ∠P \angle \mathrm{P} , to the nearest degree.\newlineAnswer:

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Q. In △OPQ,p=39 \triangle \mathrm{OPQ}, p=39 inches, o=66 o=66 inches and ∠O=81∘ \angle \mathrm{O}=81^{\circ} . Find all possible values of ∠P \angle \mathrm{P} , to the nearest degree.\newlineAnswer:
  1. Use Law of Sines: To find the possible values of ∠P\angle P, 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 angle opposite that side is the same for all three sides of the triangle. The formula is given by:\newlineasin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\newlinewhere aa, bb, and cc are the lengths of the sides, and AA, BB, and CC are the angles opposite those sides, respectively. In this case, we have side oo (opposite ∠O\angle O) and side pp (opposite ∠P\angle P), and we know ∠O\angle O.
  2. Write Equation: First, let's write down the Law of Sines for our specific triangle OPQOPQ:\newlineosin⁡(O)=psin⁡(P)\frac{o}{\sin(O)} = \frac{p}{\sin(P)}\newlineSubstituting the known values, we get:\newline66sin⁡(81°)=39sin⁡(P)\frac{66}{\sin(81°)} = \frac{39}{\sin(P)}
  3. Solve for sin⁡(P)\sin(P): Now, we solve for sin⁡(P)\sin(P):sin⁡(P)=39×sin⁡(81°)66\sin(P) = \frac{39 \times \sin(81°)}{66}We calculate sin⁡(81°)\sin(81°) using a calculator:sin⁡(81°)≈0.98768834\sin(81°) \approx 0.98768834
  4. Substitute Values: Next, we substitute the value of sin⁡(81°)\sin(81°) into our equation:\newlinesin⁡(P)=39×0.9876883466\sin(P) = \frac{39 \times 0.98768834}{66}\newlineNow, we perform the multiplication and division:\newlinesin⁡(P)≈39×0.9876883466≈0.58484848\sin(P) \approx \frac{39 \times 0.98768834}{66} \approx 0.58484848
  5. Calculate sin⁡(81°)\sin(81°): To find the angle PP, we take the inverse sine (arcsin) of sin⁡(P)\sin(P):\newlineP=arcsin⁡(0.58484848)P = \arcsin(0.58484848)\newlineUsing a calculator, we find:\newlineP≈35.68°P \approx 35.68°
  6. Find Angle P: Since we need to round to the nearest degree, ∠P\angle P is approximately:\newlineP≈36∘P \approx 36^\circ
  7. Round to Nearest Degree: However, we must consider that there could be another possible value for ∠P\angle P because the sine function is positive in both the first and second quadrants, which means there could be two different angles with the same sine value. To find the second possible value, we use the fact that the sum of the angles in a triangle is 180180 degrees. We subtract the known angles from 180180 degrees:\newline180∘−81∘−36∘=63∘180^\circ - 81^\circ - 36^\circ = 63^\circ\newlineSo the other possible value for ∠P\angle P is 63∘63^\circ.

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