Question

In \(\triangle ABC\), with usual notations, if \(\sin 2A+\sin 2B=\sin 2C\), then the type of the triangle is

• A. Isosceles triangle
• B. Equilateral triangle
• C. Right angled triangle
• D. Scalene triangle

Hint:

Use the sum to product formula and A + B = 180 minus C to show that cos A cos B = 0.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Use \(\sin 2A+\sin 2B=2\sin(A+B)\cos(A-B)\) and \(\sin 2C=2\sin C\cos C\).

Step 2: Simplify
Since \(A+B=\pi-C\), \(\sin(A+B)=\sin C\).
So \(2\sin C\cos(A-B)=2\sin C\cos C\), which gives \(\cos(A-B)=\cos C\).

Step 3: Use C = pi minus (A+B)
\(\cos C=-\cos(A+B)\), so \(\cos(A-B)+\cos(A+B)=0\).
\(2\cos A\cos B=0\).

Step 4: Conclude
So \(A=90^{\circ}\) or \(B=90^{\circ}\), and the triangle is right angled.

Final Answer:
The triangle is right angled. \[ \boxed{\text{(C) }\text{Right angled triangle}} \]