Question

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?

Answer 1

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 = $\frac{180-120}{2}=30$

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = $\frac{DC}{AC} =\frac{4}{x}$

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

                                              m∠ACD = 180-m∠ACB = 60

Hence $cos(60)=\frac{4}{x}$

          $x=\frac{4}{cos60}= 8$

Now let us consider ΔBDE, sin(∠DBE) = $\frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}$

$DE = 12sin(30) = 6cm$