A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8
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Answer:
Q=33/10-r/10 R=33−10q
Step-by-step explanation:
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Answer:
The Law of Cosine : cos C = 
Step-by-step explanation:
See the figure to understand the proof :
Let A Triangle ABC with sides a,b,c,
Draw a perpendicular on base AC of height H meet at point D
Divide base length b as AD = x -b and CD = x
By Pythagoras Theorem
In Triangle BDC And In Triangle BDA
a² = h² + x² ( 1 ) c² = h² + (x-b)²
c² = h² + x² + b² - 2xb ...(. 2)
From above eq 1 and 2
c² = (a² - x²) + x² + b² - 2xb
or, c² = a² + b² - 2xb .....(3)
Again in ΔBDC
cos C = 
Or, cos C = 
∴ x= a cos C
Now put ht value of x in eq 3
I.e, c² = a² + b² - 2ab cos C
Hence , cos C =
Proved Answer