OpenStudy (imamod):

FAN + MEDAL Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a flagpole, as shown below: Angle ADC measures 60° and angle BDC measures 30°. What is the distance between the points A and B on the flagpole?

3 years ago
OpenStudy (imamod):
3 years ago

OpenStudy (imamod):

.

3 years ago
OpenStudy (anonymous):

I got you im checking what to do

3 years ago
OpenStudy (imamod):

OK THANKS

3 years ago
OpenStudy (anonymous):

This is farely easy, i just haven't done geometry in a while. Do you know special triangles?

3 years ago
OpenStudy (imamod):

a little

3 years ago
OpenStudy (anonymous):

when you know special triangles, it will give you the value for BD

3 years ago
OpenStudy (imamod):

ok

3 years ago
OpenStudy (anonymous):

Here is a tab for the special triangle

3 years ago
OpenStudy (anonymous):
3 years ago

OpenStudy (anonymous):

do you have any answer choices/

3 years ago
OpenStudy (imamod):

yes

3 years ago
OpenStudy (imamod):

40 feet 20 feet 30 feet 10 feet

3 years ago
OpenStudy (imamod):

@ganeshie8 can you help?

3 years ago
OpenStudy (imamod):

@mathmale can you help?

3 years ago
OpenStudy (imamod):

@YanaSidlinskiy can you help?

3 years ago
OpenStudy (imamod):

@satellite73 can you help?

3 years ago
OpenStudy (yanasidlinskiy):

AC: \(\normalsize\tan(60) = \frac{AC}{10 \sqrt{3}} \implies \sqrt{3} = \frac{AC}{10 \sqrt{3}}\) \(\normalsize\ AC = 10 \sqrt{3} \times \sqrt{3} = 30\) BC: \(\normalsize\ tan(30) = \frac{BC}{10 \sqrt{3}} \implies \frac{1}{\sqrt{3}} = \frac{BC}{10 \sqrt{3}} \implies BC = 10\) \(\ AB = 30 - 10 = 20\)

3 years ago