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Question:
in a triangle ABC, sinA sinB sinC=1 √2 and cosA cosB cosC=√2 then the triangle is
Answer:

Given, sin A + sin B + sin C = 1 + √2  ..........1

cos A + cos B + cos C = √2 ........2

This is only possible, when the angles are in the form

90° , 45° and 45°

Let A = 90° , B = 45° and C = 45° 

Now, sin A + sin B + sin C = sin 90° + sin 45° + sin 45°

                                       = 1 + 1/√2 + 1/√2

                                       = 1 + 2/√2

                                       = 1 + (√2 * √2)/√2

                                       = 1 +√2

Again, cos A + cos B + cos C = cos 90° + cos 45° + cos 45°

                                           = 0 + 1/√2 + 1/√2

                                           = 2/√2

                                           = (√2 * √2)/√2

                                           = √2

Since, one of the angle is 90° , So the triangle is right angle triangle.

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