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prove pythagoras theorem using trigonometry..........................plz explain briefly

prove pythagoras theorem using trigonometry..........................plz  explain briefly
 

Grade:10

6 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
9 years ago
Hi Student

We know using trigonometry
sin\Theta = p/h
and cos\Theta = b/h
and using identity sin2\Theta + cos2\Theta =1
p2/h2 +b2/h2 =1
p2 + b2 = h2 (Hence proved)
Ravi
askIITians Faculty 69 Points
9 years ago
For a right angled triangle, with sides P(perpendicular), B(base) and H(hypotenuse);
write the trig. ratios in terms of P, B and H with any of the internal angle of the triangle.

Using the identity,
\sin^{2}\theta + \cos^{2}\theta =1;
substitute the value in terms of P,B and H. to get to the pythagorus theorem.
Saurabh Kumar
askIITians Faculty 2400 Points
9 years ago
Pythagorean Theorem, The theorem states that:

"The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs”.


This theorem is talking about the area of the squares that are built on each side of the right triangle.
245-2183_didi.gif

Accordingly, we obtain the following areas for the squares, where the green and blue squares are on the legs of the right triangle and the red square is on the hypotenuse.



area of the green square is
area of the blue square is
area of the red square is



From our theorem, we have the following relationship:

area of green square + area of blue square = area of red square or
S. Agarwal
33 Points
9 years ago
Hi !
The solutions provided by Ravi and Harsha are infact corollaries.And saurabh used and Euclidian method!!!! 
Sadly, there are no trigonometric proofs [of the Pythagorean theorem],because all of the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean theorem; because
of this theorem we say sin2 A + cos2 A = 1 etc.

For more information: Google “On the Possibility of Trigonometric Proofs of the Pythagorean Theorem”

 
Amarnath Murthy
14 Points
2 years ago
Draw a parpendicular from the vertex  C opposite diagonal on the diagonal.
Let it divide the diagonal in two parts x and y 
 c = x +y 
Using only definition of sin and cos  ( twice) 
 c = b cos A + asinA
= b.b/c + a.a/c 
→ c^2  = b^2 + a^2 
 
Amarnath Murthy
14 Points
2 years ago
Another proof 
 
Sin (A +B) = SinA CosB + CosASinB 
in right angled triangle  A + B = 90 degrees
CosB = Sin(90-B) =  SinA and SinB = Cos( 90-B)  = CosA , Sin(A+B) = Sin90 = 1
→ 1 = Sin^2 (A) + Cos^ (A)
multiply by c^2 both side 
c^2 = (cSinA)^2 + (cSinB)^2 = b^2 + a^2 
 
 

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