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Find the area of triangle ABC:

 

AB=19, ∠B= 32°.

A line is drawn from C down to AB. We call the point of intersection D. 

AD= 5, ∠ACD=10°.

 

 

I don't know how to set up the equations to find x and y. Help is much appreciate

Guest Nov 30, 2014

Best Answer 

 #1
avatar+26961 
+10

See if this helps:

triangle

(1)  10° + 32° + θ + Φ = 180°

(2) sin(32°)/CD = sin(θ)/12

(3) sin(Φ)/CD = sin(10°)/5

Three equations in three unknowns should enable you to find the angles.

 

Then use the sin rule again to find the lengths of the other sides, AC and BC.

 

Then use Herons formula to find the area:

Area = √[s(s - AB)(s - AC)(s - BC)]  where s = (AB + BC + AC)/2

Alan  Nov 30, 2014
 #1
avatar+26961 
+10
Best Answer

See if this helps:

triangle

(1)  10° + 32° + θ + Φ = 180°

(2) sin(32°)/CD = sin(θ)/12

(3) sin(Φ)/CD = sin(10°)/5

Three equations in three unknowns should enable you to find the angles.

 

Then use the sin rule again to find the lengths of the other sides, AC and BC.

 

Then use Herons formula to find the area:

Area = √[s(s - AB)(s - AC)(s - BC)]  where s = (AB + BC + AC)/2

Alan  Nov 30, 2014
 #2
avatar
+5

Thank you so much for this answer! It really helped :)

Guest Nov 30, 2014
 #3
avatar+88751 
0

Very nice, Alan

 

CPhill  Nov 30, 2014

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