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In the Triangle ADE, there is given the An∣gle* 1.1 A 30° 28', the Angle D 130° 03', and his opposite Side AE 70° 00', and 'tis required to find the Angle at E.
First by Prop. 2. Case 2. I find the Side DE, opposed to the Angle A; to be 38° 30', then proceed thus.
Fi••st find the Sum and Difference of the Sides. Then find the Difference of the Angles.* 1.2 * 1.3
Now say,
As S. ½ X. crs. DE and AE 15° 45',
To S. ½ Z. crs. EA and DE 54 15.
So is T. ½ X. VV. D and A 49 47 30",
To Tc. ½ V. at E 15° 47' 00". which doubled giveth the Angle at E 31 34, as required.