In `DeltaABC,angleA=90^(@)andtanB=(5)/(6)`, then find all other trigonometric ratios for `angleB`.

Answer

To solve the problem, we need to find all the trigonometric ratios for angle B in triangle ABC, where angle A is 90 degrees and tan B = 5/6.

### Step-by-Step Solution:

1. **Draw Triangle ABC**:
- We start by sketching triangle ABC with angle A = 90 degrees. Let angle B be one of the other angles, and angle C will be the remaining angle.

2. **Identify the Sides**:
- In triangle ABC, we know that:
- The side opposite angle B (perpendicular) = 5 (as given by tan B).
- The base (adjacent to angle B) = 6.
- Thus, we have:
- Perpendicular (opposite to angle B) = 5
- Base (adjacent to angle B) = 6

3. **Use Pythagorean Theorem to Find the Hypotenuse**:
- According to the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Perpendicular}^2 + \text{Base}^2 \]
- Substituting the values:
\[ \text{Hypotenuse}^2 = 5^2 + 6^2 = 25 + 36 = 61 \]
- Therefore, the hypotenuse is:
\[ \text{Hypotenuse} = \sqrt{61} \]

4. **Calculate Sine of B**:
- The sine of angle B is given by:
\[ \sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{\sqrt{61}} \]

5. **Calculate Cosine of B**:
- The cosine of angle B is given by:
\[ \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{6}{\sqrt{61}} \]

6. **Calculate Tangent of B**:
- We already know:
\[ \tan B = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{6} \]

7. **Calculate Cosecant of B**:
- The cosecant of angle B is the reciprocal of sine:
\[ \csc B = \frac{1}{\sin B} = \frac{\sqrt{61}}{5} \]

8. **Calculate Secant of B**:
- The secant of angle B is the reciprocal of cosine:
\[ \sec B = \frac{1}{\cos B} = \frac{\sqrt{61}}{6} \]

9. **Calculate Cotangent of B**:
- The cotangent of angle B is the reciprocal of tangent:
\[ \cot B = \frac{1}{\tan B} = \frac{6}{5} \]

### Summary of Trigonometric Ratios for Angle B:
- \(\sin B = \frac{5}{\sqrt{61}}\)
- \(\cos B = \frac{6}{\sqrt{61}}\)
- \(\tan B = \frac{5}{6}\)
- \(\csc B = \frac{\sqrt{61}}{5}\)
- \(\sec B = \frac{\sqrt{61}}{6}\)
- \(\cot B = \frac{6}{5}\)