Multiple Choice Question (MCQ)

In \triangle ABC, if \sin A\cos(\dfrac{\pi}{2}-B)=1-\sin (\dfrac{\pi}{2}-A)\cos B , then the triangle is a/n



  1. ×

    acute triangle

  2. ×

    right triangle

  3. ×

    obtuse triangle

  4. isosceles triangle

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

  1. \sin A\cos(\dfrac{\pi}{2}-B)+\sin (\dfrac{\pi}{2}-A)\cos B = 1


    \sin A(\cos\dfrac{\pi}{2} \cos B +\sin \dfrac{\pi}{2} \sin B)+ \cos B(\sin \dfrac{\pi}{2} \cos A- \cos \dfrac{\pi}{2} \sin A) = 1

    \sin A \sin B + \cos A \cos B = 1

    \cos(A-B) = 1

    A-B =0


    Therefore, the triangle is an isosceles triangle


Steven Zheng posted 2 months ago

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