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In the Triangle ADE, there is given the Side* 1.1 AE 70° 00', the Side ED 38° 30', and the An∣gle opposite thereunto at A 30° 28', and the Angle E is required.
First by Prop. 1. Case 1. I find the Angle D, opposite to AE, to be 130° 03', then proceed thus.
First find the Difference of the Angles, then find the Sum and Difference of the Sides.* 1.2 * 1.3
Now say,
As S. ½ X. crs. AE and ED 15° 45',
To S ½ Z. crs. AE and ED 54 15.
So is T. ½ X. of VV. D and A 49 47 30",* 1.4
To Tc. ½ V. at E 15° 47'. Which doubled is the Angle at E 31° 34', as was required.