In`DeltaPQR, angleR=90^(@)and"cosec"P=(13)/(5)`, then find all trigonometric ratios for `angleQ`.

Answer

To find all trigonometric ratios for angle Q in triangle PQR where angle R = 90° and cosec P = 13/5, we can follow these steps:

### Step 1: Understand the given information
We know that in triangle PQR:
- Angle R = 90°
- Cosec P = 13/5

### Step 2: Relate cosec P to sine P
Cosec P is the reciprocal of sine P:
\[ \text{cosec } P = \frac{1}{\sin P} \]
Thus,
\[ \sin P = \frac{5}{13} \]

### Step 3: Identify the sides of the triangle
In a right triangle, we can denote:
- Hypotenuse (h) = 13 (since cosec P = hypotenuse/perpendicular)
- Perpendicular (opposite to angle P) = 5
- Base (adjacent to angle P) = ?

### Step 4: Use Pythagoras theorem to find the base
According to the Pythagorean theorem:
\[ h^2 = \text{perpendicular}^2 + \text{base}^2 \]
Substituting the known values:
\[ 13^2 = 5^2 + \text{base}^2 \]
\[ 169 = 25 + \text{base}^2 \]
\[ \text{base}^2 = 169 - 25 = 144 \]
\[ \text{base} = \sqrt{144} = 12 \]

### Step 5: Identify the sides for angle Q
Now we have:
- Hypotenuse (h) = 13
- Perpendicular (opposite to angle Q) = 12
- Base (adjacent to angle Q) = 5

### Step 6: Calculate the trigonometric ratios for angle Q
1. **Sine Q**:
\[ \sin Q = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{12}{13} \]

2. **Cosine Q**:
\[ \cos Q = \frac{\text{base}}{\text{hypotenuse}} = \frac{5}{13} \]

3. **Tangent Q**:
\[ \tan Q = \frac{\text{perpendicular}}{\text{base}} = \frac{12}{5} \]

4. **Secant Q**:
\[ \sec Q = \frac{1}{\cos Q} = \frac{\text{hypotenuse}}{\text{base}} = \frac{13}{5} \]

5. **Cosecant Q**:
\[ \csc Q = \frac{1}{\sin Q} = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{13}{12} \]

6. **Cotangent Q**:
\[ \cot Q = \frac{1}{\tan Q} = \frac{\text{base}}{\text{perpendicular}} = \frac{5}{12} \]

### Final Answers:
- \(\sin Q = \frac{12}{13}\)
- \(\cos Q = \frac{5}{13}\)
- \(\tan Q = \frac{12}{5}\)
- \(\sec Q = \frac{13}{5}\)
- \(\csc Q = \frac{13}{12}\)
- \(\cot Q = \frac{5}{12}\)