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angle BAD = 135 - x, angle ACD=45-x. Using sine rule on ADB and ADC we get
AD/sinx =BD/sin(135-x) and AD/sin(45-x) = DC/sinx Since BD = DC,we get after eliminating AD
sin^2 x = sin (135 -x) sin (45 -x)
=(1/2) (cosx + sinx) (cosx - sinx)
Multiplying out 3sin^2 x =cos^2x so tanx =1 /sqrt 3 so x = 30.
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angle BAD = 135 - x, angle ACD=45-x. Using sine rule on ADB and ADC we get
AD/sinx =BD/sin(135-x) and AD/sin(45-x) = DC/sinx Since BD = DC,we get after eliminating AD
sin^2 x = sin (135 -x) sin (45 -x)
=(1/2) (cosx + sinx) (cosx - sinx)
Multiplying out 3sin^2 x =cos^2x so tanx =1 /sqrt 3 so x = 30.
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