Solve the problems pertaining to special right triangles. Leave all answers as reduced radicals.

Question

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I need help with this so i can complete my assignment. Please help?(:

Answer 1

Hi Sabrina,

 

Sin(30)=x/20, so x=20sin(30)= 20*.5 = 10              (sin = 0pp/hyp)

 

cos(30)=y/20, so y=20cos(30)= 20*√3/2=10*√3     (cos =adj/hyp)

 

Hope this helps

 

Jim

Answer 2

I think Special Triangle is the one that, its trigonometric functions can be evaluated by Geometry

 

 

 Draw a right triangle Label A, the tip vertex , C is vertical end of Vertex A , as one leg and B is the end point of horizontal leg.

 

 <C = 90°  , <A = 30° and  <B = 60°

 

   If we Extend BC equal length from C to Point D

 

    Then we have Triangle ABD as an isosceles triangle all sides equal , and all angle equals 60°

 

     Then:  Sin< A = 30°    in right triangle  ABC = --BC/---AC  = 1/2

 

                Cos<A= 30° =√( 1 - (1/2^2) = √(1 -1/4) =√3/4 = √(3) /2    

                

                Tan A = Sin 30°/ Cos 30° = ( 1/2) / (√(3) /2 = √(3)/ 3

               

                Cot A  = 1/ Tan A = √(3)

              

            Then  Sin30° = Cos 60°

                     Cos 30° = Sin 60°

                      Tan30°  =Cot 60°

           So triangle of 30,60 is a special right triangle with the above trigonometric value.

 

           Another special triangle is  90,45,45

 

           Since 2 angles are equal, then Hypotenuse with legs equal a is

 

          √( a^2 + a ^2) = √(2a^2) = a√2

     

           Sin 45° = a / ( a√2) = 1/√2 =√2 /2

 

           Cos 45° = √2/ 2

 

            Tan 45° = Cot 45° = a/a = 1

 

           Note: All trigonometric functions are evaluated by using the basic definition of

             

                  SinA = Side opposite/ Hypotenuse , and .....