To find the answers, substitute the variables (letters) in the equation for the number on that line that corresponds to that letter, then solve with the equation.

4 0

2 years ago

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?

7nadin3 [17]

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) = $\frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}$

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

7 0

3 years ago

Please help me :))) :PPPPPP

Korolek [52]

I prefer decimals with common denominators so I'm going to go ahead and convert them.

-1.25+.50 = -.75

-.75 is -3/4 aka "c" or the third option.

8 0

3 years ago

Read 2 more answers