SOLUTION: A triangle with < BAC = 20 DEG.and AB=AC. D IS such a point on AB that AD=BC.Then find out < ADC.

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Question 667490: A triangle with < BAC = 20 DEG.and AB=AC. D IS such a point on AB that AD=BC.Then find out < ADC.
Found 3 solutions by lynnlo, Edwin McCravy, AnlytcPhil:
Answer by lynnlo(4176)   (Show Source): You can put this solution on YOUR website!
ok
Answer by Edwin McCravy(20064)   (Show Source): You can put this solution on YOUR website!
Here is the triangle without the point D.



Before we put in the point D, let's chop the isosceles 
triangle into two right triangles with a median to the 
base, like this green line AE, and we will let BE = 1 unit
making BC = 2 units:



 = sec(80°),  = sec(80°), AB = sec(80°) = AC






Now we have a case of Side-angle-side with triangle ADC.

So we use the law of cosines first to find the length of CD.

CD² = AD² + AC² - 2·AD·AC·cos(20°)

CD² = 2² + sec²(80°) - 2·2·sec(80°)·cos(20°)

CD² = 4 + sec²(80°) - 4·sec(80°)·cos(20°)

CD² = 15.51754097

CD = 3.939231012

Now we turn to the law of sines:

 = 

Cross multiply:

CD·sin(ADC) = AC·sin(A)

sin(ADC) = 

sin(ADC) = 

sin(ADC) = 0.5

< ADC = 150°

Edwin
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
Here is the triangle without the point D.



Before we put in the point D, let's chop the isosceles 
triangle into two right triangles with a median to the 
base, like this green line AE, and we will let BE = 1 unit
making BC = 2 units:



 = sec(80°),  = sec(80°), AB = sec(80°) = AC






Now we have a case of Side-angle-side with triangle ADC.

So we use the law of cosines first to find the length of CD.

CD² = AD² + AC² - 2·AD·AC·cos(20°)

CD² = 2² + sec²(80°) - 2·2·sec(80°)·cos(20°)

CD² = 4 + sec²(80°) - 4·sec(80°)·cos(20°)

CD² = 15.51754097

CD = 3.939231012

Now we turn to the law of sines:

 = 

Cross multiply:

CD·sin(ADC) = AC·sin(A)

sin(ADC) = 

sin(ADC) = 

sin(ADC) = 0.5

< ADC = 150°

Edwin
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