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Asked by Renu Last Modified
Answer 
Krunal Patel
Chandan Kumar
Tutor
Kush Mudgal
In triangle ACB, angle ACB is right angle, AB = 29 unit, it is hypotenuse, BC = 21unit,
So, AC2= hypotenuse2- BC2= AB2 - BC2= 292 – 212= 202 or AC = 20 unit. angle ABC = angle β.
sin β = AC/AB = 20/29 and cos β= BC/AB = 21/29 …now
Question 1:
Sin2 β +cos2 β = (20/29)2 + (21/29)2 = 202 /292 + 212 /292 = 841/841 = 1 answer.
Question 2:
cos2 β – sin2 β = (21/29)2 –( 20/29)2 = 41/ 292 = 41/841 = .048 answer.
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Using Pythagoras’s theorem;
Third side = (29^2-21^2)^(1/2)=20
Now angle B=Cos^-1 (20/21)
I assume next part asks cos^2B+Sin^2B, this will be 1 irrespective of what is B
Second with - sign this equals to cos 2B or 2cos^2(B)-1
Just put cos B value here and get the result.
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