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? PLEASE HELP!

Answer 1

Answer:

6cm

Step-by-step explanation:

explanation attached

Answer 2

Given : ∠ACB=120°,CD=4cm

To find : distance from D to base AB

Solution :

here ,

in triangle ACD

using trigonometry ratios

$\cos 60^\circ = \frac{CD}{AC}\\\\\frac{1}{2}=\frac{4}{AC}\\\\AC = 8 cm$

AC = BC = 8 cm

BD = BC + CD

BD = 8 cm + 4 cm

BD = 12 cm

now,

$\sin 30^\circ = \frac{DE}{BD}\\\\\frac{1}{2}=\frac{DE}{12}\\\\DE = \frac{12}{2}\\\\DE = 6 cm$

hence , distance from D to base AB i.e DE = 6 cm

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