Hey! A new Challenge for you
● Try to Solve the below question in how many possible ways you can
QUESTION :
《 If angle B and angle Q are acute angles Such that SinB = SinQ , prove that angle B = angle Q ?
■ Be Brainly
✴ Happy New year !!!
Answers
1 ST METHOD
Given :
sin B = sin Q
∠B and ∠Q are acute
sin = height / hypotenuse
Let ∠ B and ∠ Q belong to Δ ABQ right ∠d at A
So : sin B = AQ / BQ
sin Q = AB / BQ
sin B = sin Q
==> AB / BQ = AQ / BQ
Cancel BQ from the denominator to get :
==> AB = AQ
Hence ∠B = ∠Q [ base ∠s of isosceles Δ ]
[ P.R.O.V.E.D ]
2 N D METHOD
Also ∠B + ∠Q = 90
∠B = 90 - ∠Q
sin² B + cos² B = 1 [ Identity of trigonometry ]
==> sin² B + cos² ( 90 - Q ) = 1
==> sin² B + sin²Q = 1
==> sin² B = 1 - sin²Q
==> sin² B = cos² Q
==> sin B = cos Q
Take the example of a Δ XBQ ,
sin B = XQ / BQ
cos Q = XB / BQ
Again : XQ / BQ = XB / BQ
==> XQ = XB
Thus ∠B = ∠Q [ base ∠s of isosceles triangle ]
3 RD METHOD [ easiest ]
sin B = sin Q
Now :
sin B = sin ( 90 - B ) [ acute angles ]
==> sin B = cos B
sin B = cos B
sin B / cos B = 1
==> tan B = 1
==> tan B = tan 45
==> ∠ B = 45
Hence ∠Q = 90 - ∠B
= 90 - 45
= 45
Hence ∠Q = ∠B = 45
4 TH METHOD [ toughest ]
sin² B + cos²B = 1
==> sin²( 90 - Q ) + cos²B = 1
==> cos²Q + cos²B = 1
By trial and error method :
Only when cos Q = cos B = 1 / √2 this will be true .
1 / 2 + 1 / 2 = 1
Hence cos Q = cos B = 1 / √2
==> cos Q = cos B = cos 45
∠ B = ∠Q = 45 [ P.R.O.V.E.D ]
If you need more methods tell in the comments
Very tired after the trial and error !!!
Hope it helps and a very happy new year sis
#JISHNU#
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1)sinB=sinQ
so: sinB=cosQ
cosB=cosQ
sin^2B+cos^2B=1
=cos^2B+cos^2B=1
=2cos^2B=1
=cos^2B=1/2
=cosB=1/√2
B=45
Q=90-45=45
B=Q
2)sinB=sinQ
sinB=sin(90-B)
=sinB=cosB
=sinB/cosB=1
=tanB=1
=B=45
Q=B=45
Hope it helps