`DeltaPQR` is right angled at Q .If cosec `P=(17)/(15)`,then what is the value of sin R?

Options

A. `(15)/(17)`
B. `(8)/(17)`
C. `(17)/(8)`
D. `(17)/(15)`

Answer

To solve the problem step by step, we will follow the trigonometric relationships in a right-angled triangle.

### Step 1: Understand the triangle
Given triangle \( \Delta PQR \) is right-angled at \( Q \). We know that:
- \( \angle P + \angle R = 90^\circ \) (since \( \angle Q = 90^\circ \))

### Step 2: Use the given cosecant value
We are given that \( \csc P = \frac{17}{15} \). Recall that:
\[ \csc P = \frac{1}{\sin P} \]
Thus, we can find \( \sin P \):
\[ \sin P = \frac{1}{\csc P} = \frac{15}{17} \]

### Step 3: Find \( \sin R \)
Using the co-function identity in a right triangle:
\[ \sin R = \cos P \]
We can find \( \cos P \) using the Pythagorean identity:
\[ \sin^2 P + \cos^2 P = 1 \]
Substituting \( \sin P \):
\[ \left(\frac{15}{17}\right)^2 + \cos^2 P = 1 \]
Calculating \( \left(\frac{15}{17}\right)^2 \):
\[ \frac{225}{289} + \cos^2 P = 1 \]
Subtracting \( \frac{225}{289} \) from both sides:
\[ \cos^2 P = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289} \]
Taking the square root:
\[ \cos P = \frac{8}{17} \]

### Step 4: Conclude with \( \sin R \)
Since \( \sin R = \cos P \):
\[ \sin R = \frac{8}{17} \]

### Final Answer
The value of \( \sin R \) is \( \frac{8}{17} \).

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