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irina irina
wrote...
Posts: 919
11 years ago
if you are solving for B, and C= 14, and A= 5

using the Pythagorean Theorem, ( a^2 + b^2 = c^2 )

how can you do this? please explain!
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wrote...
11 years ago
5^2 + b^2 = 14^2
25 + b^2 = 196
b^2 = 171
b is approximately 13
wrote...
11 years ago
5^2=25
14^2=196

196-25=171

sqrt of 171 =13.076696830622020656710945951579

B= approx. 13.1
wrote...
11 years ago
5^2+b^2=14^2
25+b^2=196
b^2=196-25=171
b=?(171)?13.1
wrote...
11 years ago
13=b
a^2 + b^2 = c^2
5^2+13^2=14^2
(5x5) + (13x13) = (14x14)
25 + 171 = 196
wrote...
11 years ago
5^2 + b^2 = 14^2
25 + b^2 = 196
b^2 = 196-25
b= 13.08
wrote...
11 years ago
|\          
   |_\
   |__\
   |___\
|B___\   C = 14
   |_____\
   |______\
   |_______\
       A = 5
Simply plug into the equation:
a^2 + b^2 = c^2
5^2 + b^2 = 14^2
25 + b^2 = 196
      -25         -25
b^2 = 171
?(b^2) = ?(171)   <=== Square Root Both Sides
b = 13.076697     <=== Approximate Answer
wrote...
11 years ago
a^2+b^2=c^2
first you plug in everything to the equation.
5^2+B^2=14^2
then you figure out what a and c are.
25+b^2=196
after that you subtract a from c.
196-25= 171
c-a=b
now you check to see if this is correct.
25+171=196
and lastly you find the square root of each # and pu tit back in the equation.
5^2+13.08^2=14
There you go. !
wrote...
11 years ago
Put your known quantities on one side of the equation:

a^2 - c^2 = -b^2 (subtracting c^2 and b^2 from both sides maintains the equality of the equation)

c^2 - a^2 = b^2

(14)^2 - (5)^2 = b^2

196 - 25 = b^2

171 = b^2

sqrt(171) = b

b = 13.077
wrote...
11 years ago
Assuming c is hypotenuse:-
c² = 5² + b²
14² = 5² + b²
b² = 14² - 5²
b² = (14 - 5)(14 + 5)
b² = 9 x 19
b² = 171
b = 13.1

PS
Use lower case letters for sides and upper case letters for angles is the usual convention.
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