Again COS PAB cot PAB= sin PAB thus the cotangent of 0o may be said to be equal to, that is to infinity. 99. It is also necessary to have a distinct conception of the limits to which the Trigonometrical Ratios tend when the angle becomes very nearly a right angle. These may be obtained from the figure, in the manner of Art. 97; or they may be deduced from the results given in Art. 97. We shall adopt the latter method. 100. Finally it is necessary to have a distinct conception of the limits to which the Trigonometrical Ratios tend when the angle becomes very nearly two right angles. These also may be obtained from the figure in the manner of Art. 97; or they may be deduced from the results given in Art. 97. We shall adopt the latter method. 64 APPLICATION OF ALGEBRAICAL SIGNS. 101. One remark may be made to prevent a possible misconception of some of the preceding results. We have put as the value of tan 90°. The student must not assume that means necessarily. If an angle is a little less than 90°, the tangent is large and positive; if the angle is a little greater than 90°, the tangent is large and negative; so that in saying that tan 90° is we must not suppose that the sign + is necessarily to be taken before ∞. So also in other cases. We have put as the value of cot 180°, whereas the student might have expected -∞. It is true that if an angle is a little less than 180°, the cotangent is negative; but if, as in a subsequent Chapter, we suppose an angle a little greater than 180°, the cotangent is positive. 102. The following table collects many of the results obtained in the present Chapter and Chapter III. EXAMPLES. IX. 1. Find the number of degrees in each angle of a regular pentagon. Find also the number of grades. 2. The number of grades in an angle is equal to twothirds of the number of degrees in the supplement of the angle: determine the angle. 3. There are as many degrees in the complement of A as in the supplement of 44: determine A. 4. Given sin A for 4. = 5 13 find the Trigonometrical Ratios 5. Given cot A=2-√3, shew that sec A= √/6+√√2, and cosec A=√√6− √2. 6. If tan A=1+√2, find cos2 A, sin2 A, and cos 24. 7. If tan A tan B=1, shew that sec A=cosec B. 9. Shew that tan 52}°= (√3+ √2) (√2−1). 10. Given sin A=m sin B, and cos A=n cos B, find sin A and sin B. 11. Given sin A=m sin B, and tan A=n tan B, find sin2A and cos2B. a2-1 13. If tan A+ sec A=a, shew that sin A a2+1 14. Given sin A+ cosec A = 2764ʊ, find sin A. (4 cos A-1) tan" A+(3-4 cos2 A)2= sec2 A. 19. Shew geometrically that sin 24 is less than 2 sin A. 20. Shew that cosec A (sec A-1)+ sin A=cot A (1-cos A)+tan A. 21. Given a=5, b=20, C=90°, find A and B. log 5='6989700, Ztan 75° 57′ = 10'6016170, 22. Given a=9'65, b=12.24, C=90°, find A and B. log 2=3010300, log 153=21846914, log 193=2·2855573, Z tan 38° 15′-98967116, Z tan 38° 16′ = 9.8969714. 23. If sin A+ cos A=a+√(1-a2), shew that either sin A or cos A is equal to a. X. Properties of Triangles. 103. The present Chapter will contain some properties of triangles which are useful for the solution of triangles. We begin with a proposition which we have already given in Art. 37, but which must be repeated in order to shew that it holds for oblique-angled triangles as well as for acute-angled triangles. We retain the notation of Art. 37. 104. In any triangle the sides are proportional to the sines of the opposite angles. Let ABC be a triangle, and from A draw AD perpendicular to the opposite side, meeting that side, or that side produced at D. If B and C are acute angles, we have from the lefthand figure If the angle C be obtuse, we have from the right-hand figure AD=AB sin B, and AD=AC sin (180°-C) |