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December 21, 2014

December 21, 2014

Posted by **joe** on Thursday, December 8, 2011 at 10:41pm.

in triangle YZX x=12cm z=16cm measure of angle Y= 130 degrees

in trianglE XYZ z= 11in. x= 7in. y= 13in.

- geometry -
**drwls**, Friday, December 9, 2011 at 7:12amUse this formula, which is attributed to Heron of Alexandria but can be traced back to Archimedes.

Area = sqrt[s(s-a)(s-b)(s-c)],

where s=(a+b+c)/2 or perimeter/2.

- help me please -
**Steve**, Friday, December 9, 2011 at 11:21amTo use that formula on the first triangle, you need to figure the length of side y

Draw the altitude h from Y to XZ, meeting XZ at P

Let angle a = <XYP

h/16 = cos a

h/12 = cos(130-a)

16cos z = 12cos(130-a)

a = 68.81°

So, side y = XZ = 16sin68.81° + 12sin61.19° = 25.432

Now you can use Heron's formula to find the area: 73.54

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