Distance between AB and DC is 17.32 cm.
If true then write 1 and if false then write 0.

Open in App

Solution

Given, trapezium ABCD, $A D = 20$ , $∠ A = 60^{\circ}$
Draw perpendicular from D on AB, which meets AB at E
Now, In $△ A D E$ ,
$sin 60^{\circ} = \frac{D E}{A D}$
$\frac{\sqrt{3}}{2} = \frac{D E}{A D}$
$D E = 10 \sqrt{3}$
$D E = 17.32 c m$

Suggest Corrections

Similar questions

1

0

2 c o s^{2} A - 1 = 1 - 2 s i n^{2} A

s i n 3 A = 3 s i n A - 4 s i n^{3} A

s i n 60^{\circ} = 2 s i n 30^{\circ} c o s 30^{\circ}

s i n (2 \times 30^{\circ}) = \frac{2 t a n 30^{\circ}}{1 + t a n^{2} 30^{\circ}}

Join BYJU'S Learning Program