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In right triangle BCD with angle D = 90 , we have BC = 8 and BD = 4. Find sin B.

 Jun 27, 2021
 #1
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Need to find side DC via Pythagorean Theorem

 

8^2 = 4^2 + (DC)^2

DC = sqrt 48

 

Sin B = opposite side / hypotenuse =   DC /hyp = sqrt 48 / 8       simplify as needed .........

 

 

 

Second method :    Cos B = adjacent side / hypot = 4/8  

  Trig identity:    sin^2 + cos^2 = 1

    then     sin = sqrt (1-cos^2) = sqrt ( 1 - 16/64) = sqrt 48 / 8   

 Jun 27, 2021
edited by ElectricPavlov  Jun 27, 2021

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