In the isosceles △abc m∠acb=120° and ad is an altitude to leg bc . what is the distance from d to base ab , if cd=4cm?

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC =

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) =

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence

Now let us consider ΔBDE, sin(∠DBE) =

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)