Question

now, give me this answer..
and don't see only..
give me the correct answer ​

Answer 1

Answer:

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Answer 2

Answer:

tan x =$\frac{5}{12}$   cos y =$\frac{4}{5}$   sin y =$\frac{3}{5}$

Step-by-step explanation:

Given that in ΔABC, sides BC= 21,  AC= 13

            if  ACD is triangle, DC = 5,  AC=13

tan x = $\frac{opposite side }{adjacent side}$ =  $\frac{DC}{AD}$  ( opposite and adjacent side of x)

cos y = $\frac{adjacent side}{hypotenuse}$ = $\frac{BD}{AB}$  ( adjacent and hypotenuse of y )

sin y = $\frac{opposite side }{hypotenuse}$ =  $\frac{AD}{AB }$( opposite and hypotenuse of y )

  now we need to find out AB, AD  to calculate values of above  

 angle at D is a right angle (90°)

 therefore Δ ADC , Δ ABD are right angle triangles

in a right angle triangle  $hypotenuse^{2} = side^{2} + side^{2}$

     in   Δ ADC =   $AC^{2} = AD^{2} + DC^{2}$

                           $13^{2} = AD^{2} + 5^{2}$

                           169 = $AD^{2}$+ 25

                           $AD^{2}$  = 169-25= 144 =$12^{2}$

      AD = 12

   in  Δ ABD =  $AB^{2} = AD^{2} +BD^{2}$

                      $AB^{2} = 12^{2} +16^{2}$   ( BD =BC -DC = 21-5=16)

                      $AB^{2} = 144+ 256$  = 400 = $20^{2}$

   AB= 20    

  tan x = $\frac{DC}{AD}$ = 5 / 12

  cos y =$\frac{BD}{AB}$ =16 / 20  = 4/5

  sin y =$\frac{AD }{AB}$ = 12/ 20 = 3/5