In `DeltaABC,angleB=90^(@)andsinA=(3)/(5)`, then find all other trigonometric ratios for `angleA`.

Answer

To solve the problem step by step, we will find all the trigonometric ratios for angle A in triangle ABC, where angle B is 90 degrees and sin A = 3/5.

### Step 1: Draw Triangle ABC
We start by drawing triangle ABC, where angle B is the right angle (90 degrees). Label the sides:
- Let AB be the side opposite angle A (perpendicular).
- Let BC be the base (adjacent to angle A).
- Let AC be the hypotenuse.

### Step 2: Identify the Given Information
We know that:
- sin A = opposite/hypotenuse = AB/AC = 3/5
From this, we can identify:
- Opposite side (AB) = 3
- Hypotenuse (AC) = 5

### Step 3: Use Pythagorean Theorem to Find the Base
According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the known values:
\[ 5^2 = 3^2 + BC^2 \]
\[ 25 = 9 + BC^2 \]
\[ BC^2 = 25 - 9 = 16 \]
Taking the square root:
\[ BC = 4 \]

### Step 4: Find Cosine of Angle A
Using the definition of cosine:
\[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4}{5} \]

### Step 5: Find Tangent of Angle A
Using the definition of tangent:
\[ \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{3}{4} \]

### Step 6: Find Cosecant of Angle A
Cosecant is the reciprocal of sine:
\[ \csc A = \frac{1}{\sin A} = \frac{5}{3} \]

### Step 7: Find Secant of Angle A
Secant is the reciprocal of cosine:
\[ \sec A = \frac{1}{\cos A} = \frac{5}{4} \]

### Step 8: Find Cotangent of Angle A
Cotangent is the reciprocal of tangent:
\[ \cot A = \frac{1}{\tan A} = \frac{BC}{AB} = \frac{4}{3} \]

### Summary of Trigonometric Ratios
- sin A = 3/5
- cos A = 4/5
- tan A = 3/4
- csc A = 5/3
- sec A = 5/4
- cot A = 4/3