PQRS is a rhombus whose diagonals are `PR=6` cm and `QS=8` cm . If O is the point of intersection of diagonals and `/_PQO=theta` then find the value of `sin theta, tan theta` and `sec theta`

Answer

To solve the problem step by step, we will analyze the rhombus PQRS and its diagonals.

### Step 1: Understand the properties of the rhombus
In a rhombus, the diagonals bisect each other at right angles (90 degrees). This means that the diagonals divide the rhombus into four right-angled triangles.

### Step 2: Determine the lengths of the segments created by the diagonals
Given:
- Diagonal PR = 6 cm
- Diagonal QS = 8 cm

Since the diagonals bisect each other:
- Length of PO = PR/2 = 6/2 = 3 cm
- Length of QO = QS/2 = 8/2 = 4 cm

### Step 3: Identify the triangle of interest
We will focus on triangle POQ, where:
- PO = 3 cm (perpendicular)
- QO = 4 cm (base)

### Step 4: Use the Pythagorean theorem to find the hypotenuse PQ
Using the Pythagorean theorem:
\[ PQ^2 = PO^2 + QO^2 \]
\[ PQ^2 = 3^2 + 4^2 = 9 + 16 = 25 \]
\[ PQ = \sqrt{25} = 5 \text{ cm} \]

### Step 5: Calculate sin(theta)
In triangle POQ, we know:
- Opposite side (perpendicular) = QO = 4 cm
- Hypotenuse = PQ = 5 cm

Thus,
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QO}{PQ} = \frac{4}{5} \]

### Step 6: Calculate tan(theta)
In triangle POQ, we know:
- Opposite side (perpendicular) = QO = 4 cm
- Adjacent side (base) = PO = 3 cm

Thus,
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QO}{PO} = \frac{4}{3} \]

### Step 7: Calculate sec(theta)
Secant is the reciprocal of cosine. First, we need to find cos(theta):
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PO}{PQ} = \frac{3}{5} \]
Thus,
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{5}{3} \]

### Final Answers
- \(\sin \theta = \frac{4}{5}\)
- \(\tan \theta = \frac{4}{3}\)
- \(\sec \theta = \frac{5}{3}\)