The length of AD is

Question

The length of AD is

Toppr Admin · Accepted Answer

In -x25B3-ABCAC-BC-x2220-ABC-x2220-CAB-x -xA0- -xA0- -xA0- -xA0- -xA0-Isosceles triangle property-Now- -x2220-ABC-x2220-CAB-x2220-ACD -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -Exterior angle property-x-x-60x-30-x2218-Now- In -x25B3-ABD-ADAB-sin30AD100-12AD-50cm

The length of $A D$ is

In $△ A B C$
$A C = B C$
$∠ A B C = ∠ C A B = x$ ...(Isosceles triangle property)
Now, $∠ A B C + ∠ C A B = ∠ A C D$ ...(Exterior angle property)
$x + x = 60$
$x = 30^{\circ}$
Now, In $△ A B D$ ,
$\frac{A D}{A B} = sin 30$

$\frac{A D}{100} = \frac{1}{2}$

$A D = 50$ cm

The length of AD is

In △ABCAC=BC∠ABC=∠CAB=x ...(Isosceles triangle property)Now, ∠ABC+∠CAB=∠ACD ...(Exterior angle property)x+x=60x=30∘Now, In △ABD,ADAB=sin30AD100=12AD=50cm

The length of $A D$ is

In $△ A B C$
$A C = B C$
$∠ A B C = ∠ C A B = x$ ...(Isosceles triangle property)
Now, $∠ A B C + ∠ C A B = ∠ A C D$ ...(Exterior angle property)
$x + x = 60$
$x = 30^{\circ}$
Now, In $△ A B D$ ,
$\frac{A D}{A B} = sin 30$

$\frac{A D}{100} = \frac{1}{2}$

$A D = 50$ cm