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In the Triangle ADE, there is given the Side* 1.1 AE 70° 00', the Side DE 38° 30', and the Angle A 30° 28', the Side AD is required.
First by Case 1. Prop. 1. I find the Angle at D to be 130° 03', and then proceed thus
First take the Sum and Difference of the two Angles; then also find the Difference of the two Sides given, and then work as followeth.* 1.2 * 1.3
Now say,
As S. ½ X. VV. D and A, 49° 47' 30",
To S. ½ Z. VV. D and 〈…〉〈…〉, 80 15 30.
So is T. ½ X. crs. AE and ED, 15 45 00,
To T. ½ cr. AD. 20° 00' 00": which doubled giveth the Side AD, 40 00 00, as was required.