AD

Question

AD

Toppr Admin · Accepted Answer

Given- AB-12- -x2220-ACB-30-x2218-x2220-ACB-x2220-ABC-x2220-BAC-18030-90-x2220-BAC-180-x2220-BAC-60-x2218-Now- In -x25B3-ABDNow- cos30-BH-ADAB12-AD12AD-6

AD

Given: AB=12, ∠ACB=30∘∠ACB+∠ABC+∠BAC=18030+90+∠BAC=180∠BAC=60∘Now, In △ABDNow, cos30=BH=ADAB12=AD12AD=6

Given: $A B = 12$ , $∠ A C B = 30^{\circ}$
$∠ A C B + ∠ A B C + ∠ B A C = 180$
$30 + 90 + ∠ B A C = 180$
$∠ B A C = 60^{\circ}$
Now, In $△ A B D$
Now, $cos 30 = \frac{B}{H} = \frac{A D}{A B}$
$\frac{1}{2} = \frac{A D}{12}$
$A D = 6$