Math, asked by sashankpavuluri20005, 5 months ago

1. If tan A = 1, show that sin A cos A = 1/2 (Using Pyth )​

Answers

Answered by ItzArchimedes
9

Solution :-

Given ,

  • tanA = 1

We need to show that ,

  • sinA cosA = 1/2 [ Using Pythagoras theorem ]

If tanA = 1 height of triangle is 1 & base of triangle is 1 .

Now , finding hypotenuse using Pythagoras theorem .

Hyp² = Base² + Height²

→ Hyp² = 1² + 1²

→ Hyp² = 1 + 1

→ Hyp² = 2

Hypotenuse = 2

sinA = opposite/Hypotenuse

cosA = Adjacent/Hypotenuse

♦ sinA = 1/√2

♦ cosA = 1/√2

Now finding, sinA cosA

⇒ 1/√2 × 1/√2

⇒ 1/(√2)²

sinA cosA = 1/2

Hence , sinA cosA = 1/2


Cynefin: Awesome :)
MisterIncredible: Brilliant bro w(°o°)w
amitkumar44481: Great :-)
Answered by Anonymous
128

♣ Qᴜᴇꜱᴛɪᴏɴ :

  • If tan A = 1, show that sin A cos A = \sf{\dfrac{1}{2}}  (Using Pythagoras Theorem )​

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♣ ᴛᴏ ᴘʀᴏᴠᴇ :

sin A cos A = \sf{\dfrac{1}{2}}  (Using Pythagoras Theorem )​

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♣ ᴀɴꜱᴡᴇʀ :

\setlength{\unitlength}{.4in} \begin{picture}(7,5)(0,0) \linethickness{1pt} \put(0,0){\line(1,0){4}} \put(4,0){\line(0,1){3}} \put(0,0){\line(4,3){4}} \put(2,-.25){\makebox(0,-1){\sf{Base}}} \put(4.25,1.5){\makebox(2,0){\sf{Height}}} \put(2,2){\makebox(-2,0){\bf{Hypotenuse}}} \end{picture}

We know :

\large\boxed{\sf{tan A = \dfrac{Height}{Base}}}

Also :

\boxed{\sf{sinA=\dfrac{Height}{Hypotenuse}}}

\boxed{\sf{cosA=\dfrac{Base}{Hypotenuse}}}

\setlength{\unitlength}{.4in} \begin{picture}(7,5)(0,0) \linethickness{1pt} \put(0,0){\line(1,0){4}} \put(4,0){\line(0,1){3}} \put(0,0){\line(4,3){4}} \put(2,-.25){\makebox(0,-1){\sf{1}}} \put(4.25,1.5){\makebox(0.5,0){\sf{1}}} \put(2,2){\makebox(-2,0){\bf{Hypotenuse}}} \end{picture}

If tanA = 1

  • Base = 1
  • Height = 1

According to Pythagoras Theorem :

Hypotenuse² = Height² + Base²

According To Question :

Hypotenuse² = 1² + 1²

Hypotenuse² = 1 + 1

Hypotenuse² = 2

Hypotenuse = √2

Now we need to Prove sin A cos A = \sf{\dfrac{1}{2}}

→ Taking L.H.S

sin A cos A

=\sf{\dfrac{Height}{Hypotenuse}\:\times \:\dfrac{Base}{Hypotenuse}}

=\sf{\dfrac{1}{\sqrt{2}}\:\times\:\dfrac{1}{\sqrt{2}}}

=\sf{\dfrac{1}{\sqrt{2}^2}}

=\sf{\dfrac{1}{2}}

= R.H.S

★  Hence Proved

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♣ ᴏᴛʜᴇʀ ᴛʀɪɢᴏɴᴏᴍᴇᴛʀɪᴄ ɪᴅᴇɴᴛɪᴛɪᴇꜱ  :

  • sin(θ) = Height / Hypotenuse
  • cos(θ) = Base / Hypotenuse
  • tan(θ) = Height / Base
  • cosec(θ) =  Hypotenuse  / Height
  • sec(θ) =  Hypotenuse  / Base  
  • tan(θ) =  Base / Height  

Attachments:

Cynefin: Awesome。◕‿◕。
MisterIncredible: Perfect
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