AB is 17-32 cmIf true then enter 1 and false then enter 0

Question

Toppr Admin · Accepted Answer

In -x25B3-ABCtan60-ABBCAB-x221A-3BCLet BC-x- AB-x221A-3xIn -x25B3-ABD-tan30-ABBDBD-AB-x221A-3BC-CD-AB-x221A-3x-20-x-x221A-3-x221A-3-x-20-3xx-10Thus- BC - 10- AB - 17-32 cm

AB is 17.32 cm
If true then enter $1$ and if false then enter $0$

In $△ A B C$
$tan 60 = \frac{A B}{B C}$
$A B = \sqrt{3} B C$
Let $B C = x$ , $A B = \sqrt{3} x$
In $△ A B D$ ,
$tan 30 = \frac{A B}{B D}$
$B D = A B \sqrt{3}$
$B C + C D = A B \sqrt{3}$
$x + 20 = x \sqrt{3} (\sqrt{3})$
$x + 20 = 3 x$
$x = 10$
Thus, BC = 10, AB = 17.32 cm

AB is 17.32 cmIf true then enter 1 and if false then enter 0

In △ABCtan60=ABBCAB=√3BCLet BC=x, AB=√3xIn △ABD,tan30=ABBDBD=AB√3BC+CD=AB√3x+20=x√3(√3)x+20=3xx=10Thus, BC = 10, AB = 17.32 cm

AB is 17.32 cm
If true then enter $1$ and if false then enter $0$

In $△ A B C$
$tan 60 = \frac{A B}{B C}$
$A B = \sqrt{3} B C$
Let $B C = x$ , $A B = \sqrt{3} x$
In $△ A B D$ ,
$tan 30 = \frac{A B}{B D}$
$B D = A B \sqrt{3}$
$B C + C D = A B \sqrt{3}$
$x + 20 = x \sqrt{3} (\sqrt{3})$
$x + 20 = 3 x$
$x = 10$
Thus, BC = 10, AB = 17.32 cm