Ab is 47-32 cmIf true then enter 1 and false then enter 0

Question

Toppr Admin · Accepted Answer

In -x25B3-ABCtan60-ABBCAB-x221A-3BC-xLet BC-x-AB-x221A-3xIn -x25B3-ABD-tan45-ABBDBD-ABBC-CD-ABx-20-x-x221A-3x-20-x221A-3-x2212-1x-27-32Thus- AB-27-32 cm and BC-27-32-x221A-3-47-32

Ab is 47.32 cm
If true then enter $1$ and if false then enter $0$

In $△ A B C$
$tan 60 = \frac{A B}{B C}$
$A B = \sqrt{3} B C = x$
Let $B C = x, A B = \sqrt{3} x$
In $△ A B D$ ,
$tan 45 = \frac{A B}{B D}$
$B D = A B$
$B C + C D = A B$
$x + 20 = x \sqrt{3}$
$x = \frac{20}{\sqrt{3} - 1}$
$x = 27.32$
Thus, $A B = 27.32$ cm and $B C = 27.32 \sqrt{3} = 47.32$

Ab is 47.32 cmIf true then enter 1 and if false then enter 0

In △ABCtan60=ABBCAB=√3BC=xLet BC=x,AB=√3xIn △ABD,tan45=ABBDBD=ABBC+CD=ABx+20=x√3x=20√3−1x=27.32Thus, AB=27.32 cm and BC=27.32√3=47.32

Ab is 47.32 cm
If true then enter $1$ and if false then enter $0$

In $△ A B C$
$tan 60 = \frac{A B}{B C}$
$A B = \sqrt{3} B C = x$
Let $B C = x, A B = \sqrt{3} x$
In $△ A B D$ ,
$tan 45 = \frac{A B}{B D}$
$B D = A B$
$B C + C D = A B$
$x + 20 = x \sqrt{3}$
$x = \frac{20}{\sqrt{3} - 1}$
$x = 27.32$
Thus, $A B = 27.32$ cm and $B C = 27.32 \sqrt{3} = 47.32$