15 In $$ \Delta x Y Z, Y=14 \mathrm{~cm} $$. Solve the triangles completely, giving the two possible solutions. 14 Calculate the values of angles $$ A $$ and $$ C $$ of $$ riangle A B C $$, where $$ b=14.35 \mathrm{~cm}, a=7.82 \mathrm{~cm} $$ and $$ B=115.6^{\circ} \cdot(W A E C) $$ (15) In $$ riangle P Q R, P \widehat{Q R}=84^{\circ}, \widehat{Q P R}=43^{\circ} $$ and

Question

15 In $$ \Delta x Y Z, Y=14 \mathrm{~cm} $$. Solve the triangles completely, giving the two possible solutions. 14 Calculate the values of angles $$ A $$ and $$ C $$ of $$ 	riangle A B C $$, where $$ b=14.35 \mathrm{~cm}, a=7.82 \mathrm{~cm} $$ and $$ B=115.6^{\circ} \cdot(W A E C) $$ (15) In $$ 	riangle P Q R, P \widehat{Q R}=84^{\circ}, \widehat{Q P R}=43^{\circ} $$ and

UpStudy AI · Accepted Answer

Sure, let's tackle the problems you've provided step by step. It appears there are two distinct problems labeled as 14 and 15. We'll address each one separately.

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### **Problem 14**

**Given:**
- Triangle $$ \triangle ABC $$
- Side $$ b = 14.35 $$ cm (opposite angle $$ B $$)
- Side $$ a = 7.82 $$ cm (opposite angle $$ A $$)
- Angle $$ B = 115.6^\circ $$

**Find:**
- Angles $$ A $$ and $$ C $$

**Solution:**

To find angles $$ A $$ and $$ C $$ in $$ \triangle ABC $$, we can use the **Law of Sines**, which states:

$$
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
$$

1. **Find Angle $$ A $$:**

Using the Law of Sines:
   
   $$
   \frac{a}{\sin A} = \frac{b}{\sin B}
   $$
   
   Solve for $$ \sin A $$:
   
   $$
   \sin A = \frac{a \cdot \sin B}{b}
   $$
   
   Plugging in the known values:
   
   $$
   \sin A = \frac{7.82 \cdot \sin(115.6^\circ)}{14.35}
   $$
   
   Calculate $$ \sin(115.6^\circ) $$:
   
   $$
   \sin(115.6^\circ) \approx 0.9018
   $$
   
   Now, compute $$ \sin A $$:
   
   $$
   \sin A = \frac{7.82 \times 0.9018}{14.35} \approx \frac{7.058}{14.35} \approx 0.4928
   $$
   
   Determine angle $$ A $$:
   
   $$
   A = \arcsin(0.4928) \approx 29.5^\circ
   $$
   
2. **Find Angle $$ C $$:**

The sum of angles in any triangle is $$ 180^\circ $$:
   
   $$
   A + B + C = 180^\circ
   $$
   
   Plugging in the known values:
   
   $$
   29.5^\circ + 115.6^\circ + C = 180^\circ
   $$
   
   Solve for $$ C $$:
   
   $$
   C = 180^\circ - 115.6^\circ - 29.5^\circ = 34.9^\circ
   $$
   
**Answer for Problem 14:**
- $$ \angle A \approx 29.5^\circ $$
- $$ \angle C \approx 34.9^\circ $$

---

### **Problem 15**

**Given:**
- Triangle $$ \triangle PQR $$
- Angle $$ \angle Q R = 84^\circ $$
- Angle $$ \angle Q P R = 43^\circ $$

**Find:**
- The remaining angle(s) or side(s) (The problem statement seems incomplete.)

**Solution:**

From the information provided, we can determine the third angle in $$ \triangle PQR $$ since the sum of angles in any triangle is $$ 180^\circ $$.

1. **Find the Third Angle $$ \angle PQR $$:**

$$
   \angle PQR = 180^\circ - \angle QPR - \angle Q R = 180^\circ - 43^\circ - 84^\circ = 53^\circ
   $$
   
2. **Determine Sides (If Needed):**

Since only angles are provided without any side lengths, we cannot determine the lengths of the sides without additional information. If side lengths are provided, we can use the Law of Sines or Law of Cosines to find the missing sides.

**Answer for Problem 15:**
- $$ \angle PQR = 53^\circ $$
  
*Note:* To fully solve **Problem 15**, additional information such as the length of at least one side is required. If you provide that, I can assist you further!

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If you have more details or additional problems you'd like assistance with, feel free to share!
**Problem 14:** - $$ \angle A \approx 29.5^\circ $$ - $$ \angle C \approx 34.9^\circ $$ **Problem 15:** - $$ \angle PQR = 53^\circ $$ *Note:* Additional information is needed to determine the sides of $$ \triangle PQR $$.

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