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In /_\NOP,n=300cm,p=100cm and /_P=45^(@). Find all possible values of /_N, to the nearest degree. Answer:

In △NOP,n=300 cm,p=100 cm \triangle \mathrm{NOP}, n=300 \mathrm{~cm}, p=100 \mathrm{~cm} and ∠P=45∘ \angle \mathrm{P}=45^{\circ} . Find all possible values of ∠N \angle \mathrm{N} , to the nearest degree.\newlineAnswer:

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Q. In △NOP,n=300 cm,p=100 cm \triangle \mathrm{NOP}, n=300 \mathrm{~cm}, p=100 \mathrm{~cm} and ∠P=45∘ \angle \mathrm{P}=45^{\circ} . Find all possible values of ∠N \angle \mathrm{N} , to the nearest degree.\newlineAnswer:
  1. Apply Law of Sines: To find the possible values of angle NN, 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 sides of the triangle. The formula is asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}, where aa, bb, and cc are the lengths of the sides, and AA, BB, and CC are the opposite angles.
  2. Write Law of Sines: First, let's write down the Law of Sines for our triangle NOP: 300sin⁡(N)=100sin⁡(45°)\frac{300}{\sin(N)} = \frac{100}{\sin(45°)}
  3. Solve for sin⁡(N)\sin(N): Now, we can solve for sin⁡(N)\sin(N):sin⁡(N)=300×sin⁡(45°)100\sin(N) = \frac{300 \times \sin(45°)}{100}
  4. Calculate sin⁡(45°)\sin(45°): Calculate sin⁡(45°)\sin(45°), which is 2/2\sqrt{2}/2 or approximately 0.70710.7071:\newlinesin⁡(N)=300×0.7071100\sin(N) = \frac{300 \times 0.7071}{100}
  5. Perform Multiplication and Division: Perform the multiplication and division to find sin⁡(N)\sin(N):sin⁡(N)≈300×0.7071/100\sin(N) \approx 300 \times 0.7071 / 100sin⁡(N)≈212.13/100\sin(N) \approx 212.13 / 100sin⁡(N)≈2.1213\sin(N) \approx 2.1213

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