Maths-
General
Easy

Question

In triangle ABC, the measure of angle C is 90 to the power of ring operator end exponent. If s i n invisible function application A equals fraction numerator 3 over denominator 5 end fraction, what is c o s invisible function application B ?

  1. fraction numerator 3 over denominator 5 end fraction    
  2. fraction numerator 4 over denominator 5 end fraction    
  3. fraction numerator 5 over denominator 4 end fraction    
  4. fraction numerator 5 over denominator 3 end fraction    

The correct answer is: fraction numerator 3 over denominator 5 end fraction



    In a right-angled triangle, sin invisible function application angle equals fraction numerator h e i g h t over denominator h y p o t e n u s e end fraction and cos invisible function application angle equals fraction numerator b a s e over denominator h y p o t e n u s e end fraction , the height and base are assigned to w.r.t. the selection of the angle.
    For example, in case of the given diagram, sin invisible function application A equals fraction numerator stack B C with minus on top over denominator stack A B with minus on top end fraction , since stack B C with minus on top is the height w.r.t. angle A and cos invisible function application B equals fraction numerator stack B C blank with minus on top over denominator stack A B with minus on top end fraction , since stack B C with minus on top is the base w.r.t. angle B.
    Explanations:
    Step 1 of 3:
    We are given that in triangle A B C, angle C equals 90 degree and sin invisible function application A equals fraction numerator 3 over denominator 5 end fraction .
    So, as per the formula, we have height open parentheses stack B C with minus on top close parentheses equals 3 unit and hypotenuse open parentheses stack A B with minus on top close parentheses equals 5 unit w.r.t. angle A.
    Step 2 of 3:
    Now, w.r.t. angle B, the base is stack B C with minus on top and hypotenuse is stack A B with minus on top.
    Step 3 of 3:
    Hence, cos invisible function application B equals fraction numerator b a s e over denominator h y p o t e n u s e end fraction equals fraction numerator stack B C with minus on top over denominator stack A B with minus on top end fraction equals fraction numerator 3 over denominator 5 end fraction
    Final Answer:
    cos invisible function application B is— fraction numerator 3 over denominator 5 end fraction .

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