Question

Given: ΔАВС, m∠ACB = 90°. CD⊥ AB. m∠ACD = 60°. BC = 6 cm. Find: АD
PLEASE HELP ME :(

Answer 1

Answer:

AD=

$3 \sqrt{5}$

Step-by-step explanation:

Seg CD perpendicular to line AB i.e.AD=DB

In traingle ABC, taking pythagoras theorem

(AB)^2 = (AC)^2 + (BC)^2

i.e (2AD)^2= (AC)^2 + (6)^2

(4AD)^2 =(AC)^2 + 36

4 ( AD^2)- AC^2 = 36

AD^2 AC^2 = 36/4

AD^2 - AC^2 = 9

Now In triangle ADC, taking pythagoras theorem

AC^2 = AD^2 + DC^2

i.e AD^2+DC^2 = AC^2

AD^2 -AC^2= -DC^2

9 = -DC^2 ( Because AD^2- AC^2 = 9)

+-3 = -DC (Taking square root on both side)

-3 = -DC

3 = DC

Now in triangle CDB , Taking pythagoras theorem

BD^2= CD^2 + BC^2

BD^2= 3^2 + 6^2

BD^2= 9+36

BD^2= 45

BD^2= 9*5

BD = 3 root 5

i.e AD = BD = 3 root 5

$3 \sqrt5]{?}$