Synopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1898 Excerpt: ...intersect at right angles. Ex. 113. Given two perpendiculars, AB and CD, intersecting in 0, and a straight line intersecting these perpendiculars in E and F; to construct a square, one of whose angles shall coincide with one of the right angles at 0, and the vertex of the opposite angle of the square Bhall lie in EF. (Two solutions.) Proposition XXXIII. Problem. 283. Two sides of a triangle and the angle opposite one of them being given, to construct the triangle. Case I. If the side opposite the given angle is less than the other given side. Let b be greater than a, and A the given angle. To construct the triangle. Construction. Construct Z DAE--to the given Z. A. § 276 On AD take AB = b. From B as a centre, with a radius equal to a, describe an arc intersecting the line AE at Cand C. Draw BC and BC. Then both the A ABC&nd ABC,X fulfil the conditions, and hence we have two constructions. This is called the ambiguous case. Discussion, If the side a is equal _ to the J. BBT, the arc described from B will touch AE, and there will be but one construction, the right triangle ABU. If the given side a is less than the _L from B, the arc described from B will not intersect or touch AE, and--hence the problem is impossible, V If the Z A is right or obtuse, the problem is impossible; for the side opposite a right or obtuse angle is the greatest side. § 159 Case II. If a is equal to b. If the Z A is acute, and a--b, the arc described from B as a centre, and with a radius equal to a, will cut the line AE at the points A and C. B,P There is therefore but one solution: the b/' isosceles A ABC. X X E Discussion. If the Z Am right or obtuse, the problem is impossible; for equal sides of a A have equal A opposite them, and a A cannot have two right A or two o...
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