Questions: In Δ NOP, p=5.1 inches, n=6.6 inches and ∠O=18°. Find ∠P, to the nearest 10th of a degree.

Transcript text: In $\Delta \mathrm{NOP}, p=5.1$ inches, $n=6.6$ inches and $\angle \mathrm{O}=18^{\circ}$. Find $\angle \mathrm{P}$, to the nearest 1oth of an degree.

Solution

Solution Steps

To find $\angle \mathrm{P}$ in $\Delta \mathrm{NOP}$, we can use the Law of Sines, which states that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Here, we know $p = 5.1$, $n = 6.6$, and $\angle \mathrm{O} = 18^\circ$. We can set up the equation using the known values and solve for $\sin \angle \mathrm{P}$, then use the inverse sine function to find $\angle \mathrm{P}$.

Step 1: Given Values

We are given the following values in triangle $ \Delta \mathrm{NOP} $:

$ p = 5.1 $ inches
$ n = 6.6 $ inches
$ \angle O = 18^\circ $

Step 2: Convert Angle to Radians

To use the sine function, we convert $ \angle O $ from degrees to radians: \[ \angle O_{\text{rad}} = \frac{18 \times \pi}{180} \approx 0.3142 \]

Step 3: Apply the Law of Sines

Using the Law of Sines, we set up the equation: \[ \frac{n}{\sin O} = \frac{p}{\sin P} \] From this, we can express $ \sin P $ as: \[ \sin P = \frac{p \cdot \sin O}{n} \]

Step 4: Calculate $ \sin P $

Substituting the known values: \[ \sin P = \frac{5.1 \cdot \sin(0.3142)}{6.6} \approx 0.2388 \]

Step 5: Find $ \angle P $

Now, we find $ \angle P $ using the inverse sine function: \[ \angle P = \arcsin(0.2388) \approx 13.8149^\circ \]

Step 6: Round to the Nearest Tenth

Rounding $ \angle P $ to the nearest tenth of a degree gives: \[ \angle P \approx 13.8^\circ \]