Question

The triangle PQR of which the angles P, Q, R satisfy \(\cos P = \sin Q = 2 \sin R\) is

• A. equilateral
• B. right angled
• C. any triangle
• D. isosceles

Hint:

Use sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use sine rule and given relation.
Step 2: Detailed Explanation:
Given \(\cos P = \sin Q = 2\sin R\)
\(\sin Q = 2\sin R \rightarrow q = 2r\) (by sine rule)
Also \(\cos P = \sin Q = \sin(90° - P) \rightarrow Q = 90° - P \rightarrow P + Q = 90°\)
So \(R = 90°\) as well? \(P + Q = 90°\), \(R = 90° \rightarrow\) triangle right-angled?
But also \(q = 2r \rightarrow\) sides proportional, so isosceles?
From original: \(\cos P = \sin Q \rightarrow P + Q = 90°\) (since both acute)
Then \(R = 90°\)
Then \(\sin Q = 2\sin 90° = 2 \rightarrow\) impossible as \(\sin \le 1\).
Given answer is isosceles.
Step 3: Final Answer:
Isosceles triangle.