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?
1 answer:
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) = 

You might be interested in
303.875303.875303.875303.875303.875303.875303.875
The probability of selecting a solid chocolate piece followed by two cherry-filled chocolates, would be 27.387 or as rounded would be 27.4
Answer:168
Step-by-step explanation:
The second option is the answer
The answer is 1,342 cm squared. This problem is fairly simple, but ok.