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

Answer

To find all the trigonometric ratios for angle A in triangle ABC, where angle B is 90 degrees and sin A = 3/5, we can follow these steps:

### Step 1: Understand the sine definition
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. Here, we have:
- sin A = opposite/hypotenuse = 3/5.

### Step 2: Assign lengths to the sides
Let the length of the side opposite angle A be 3 units and the length of the hypotenuse be 5 units. We can denote:
- Opposite side (to angle A) = 3,
- Hypotenuse = 5.

### Step 3: Use the Pythagorean theorem to find the adjacent side
In a right triangle, the Pythagorean theorem states that:
\[ a^2 + b^2 = c^2 \]
where \( c \) is the hypotenuse, \( a \) is the opposite side, and \( b \) is the adjacent side. Here, we have:
- \( a = 3 \) (opposite),
- \( c = 5 \) (hypotenuse).

We need to find \( b \) (adjacent side):
\[ 3^2 + b^2 = 5^2 \]
\[ 9 + b^2 = 25 \]
\[ b^2 = 25 - 9 \]
\[ b^2 = 16 \]
\[ b = 4 \]

### Step 4: Calculate other trigonometric ratios
Now that we have all sides of the triangle:
- Opposite = 3,
- Adjacent = 4,
- Hypotenuse = 5.

We can find the other trigonometric ratios:

1. **Cosine (cos A)**:
\[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5} \]

2. **Tangent (tan A)**:
\[ \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \]

3. **Cosecant (cosec A)**:
\[ \csc A = \frac{1}{\sin A} = \frac{5}{3} \]

4. **Secant (sec A)**:
\[ \sec A = \frac{1}{\cos A} = \frac{5}{4} \]

5. **Cotangent (cot A)**:
\[ \cot A = \frac{1}{\tan A} = \frac{4}{3} \]

### Summary of Trigonometric Ratios for Angle A:
- sin A = 3/5
- cos A = 4/5
- tan A = 3/4
- cosec A = 5/3
- sec A = 5/4
- cot A = 4/3