Plane Trigonometry

Solution of Triangles

718 Right-angled triangles are solved by formula 𝑐²=𝑎²+𝑏² 719 {𝑎=𝑐sin𝐴𝑏=𝑐cos𝐴𝑎=𝑏tan𝐴 ⋯

Scalene Triangles

720

Case I

The equation 𝑎sin𝐴=𝑏sin𝐵701 will determine any one of the four quantities 𝐴, 𝐵, 𝑎, 𝑏 when the remaining three are known.

The Ambiguous Case

721 When, in Case I, two sides and an acute angle opposite to one of them are given, we have,, from the figure, sin𝐶=𝑐sin𝐴𝑎 Then 𝐶 and 180−𝐶 are the vlues of 𝐶 and 𝐶′, by (622). Also 𝑏=𝑐cos𝐴±𝑎²−𝑐²sin²𝐴 because 𝑏=𝐴𝐷±𝐷𝐶 722 When an angle 𝐵 is to be determined from the equation sin𝐵=𝑏𝑎sin𝐴 and 𝑏𝑎 is a small fraction; the circular measure of 𝐵 may be approximated to by putting sin(𝐵+𝐶) for sin𝐴, and using theorem (796). 723

Case II

When two sides 𝑏, 𝑐 and the included angle 𝐴 are known, the third side 𝑎 is given by the formula 𝑎²=𝑏²+𝑐²−2𝑏𝑐cos𝐴702 when logarithms are not used. 724 Otherwise, employ the following formula with logarithms, tan𝐵−𝐶2=𝑏−𝑐𝑏+𝑐cot𝐴2 725 Obtained from 𝑏−𝑐𝑏+𝑐=sin𝐵−sin𝐶sin𝐵+sin𝐶(701), and then applying (670) and (671).
𝐵+𝐶2 having been found from the above equation, and 𝐵+𝐶2 being equal to 90°−𝐴2, we have 𝐵=𝐵+𝐶2+𝐵−𝐶2, 𝐶=𝐵+𝐶2−𝐵−𝐶2 𝐵 and 𝐶 having been determined 𝑎 can be found by Case I. 726 If the logarithms of 𝑏 and 𝑐 are known, the trouble of taking out log(𝑏−𝑐) and log(𝑏+𝑐) may be avoided by employing the subsidiary angle 𝜃=tan⁻¹𝑏𝑐, and the formula 727 tan12(𝐵−𝐶)=tan𝜃−𝜋4cot𝐴2655 728 Or else the subsidiary angle 𝜃=cos⁻¹𝑐𝑏, and the formula tan12(𝐵−𝐶)=tan²𝜃2cot𝐴2643 729 𝑎=(𝑏+𝑐)sin𝐴2cos12(𝐵−𝐶)From the figure in 960, by drawing a perpendicular from 𝐵 to 𝐸𝐶 produced. 730 If 𝑎 be required in terms of 𝑏, 𝑐 and 𝐴 alone, and in a form adapted to logarithmic computation, employ the subsidiary angle 𝜃=sin⁻¹4𝑏𝑐(𝑏+𝑐)²cos²𝐴2 and the formula 𝑎=(𝑏+𝑐)cos𝜃702, 637

Case III

When the three sides are known, the angles may be found without employing logarithms, from the formula 731 cos𝐴=𝑏²+𝑐²−𝑎²2𝑏𝑐703 732 If logarithms are to be used, take the formula for sin𝐴2, cos𝐴2, tan𝐴2, (704), and (705).

Quadrilateral Inscribed in a Circle

733 cos𝐵=𝑎²+𝑏²−𝑐²−𝑑²2(𝑎𝑏+𝑐𝑑) From 𝐴𝐶²=𝑎²+𝑏²−2𝑎𝑏cos𝐵=𝑐²+𝑑²+2𝑐𝑑cos𝐵, by (702), and 𝐵+𝐷=180°. 734 sin𝐵=2𝑄𝑎𝑏+𝑐𝑑613, 733 735 𝑄=(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)(𝑠−𝑑)=area of 𝐴𝐵𝐶𝐷 and 𝑠=12(𝑎+𝑏+𝑐+𝑑) Area=12𝑎𝑏sin𝐵+12𝑐𝑑sin𝐵; substitute sin𝐵 from last. 736 𝐴𝐶²=(𝑎𝑐+𝑏𝑑)(𝑎𝑑+𝑏𝑐)(𝑎𝑏+𝑐𝑑)702, 733 737 Radius of circumscribed circle =14𝑄(𝑎𝑏+𝑐𝑑)(𝑎𝑐+𝑏𝑑)(𝑎𝑑+𝑏𝑐)713, 734, 736 738 if 𝐴𝐷 bizect the side of the triangle 𝐴𝐵𝐶 in 𝐷, tan𝐵𝐷𝐴=4△𝑏²−𝑐² 739 cot𝐵𝐴𝐷=2cot𝐴+cot𝐵 740 𝐴𝐷²=14(𝑏²+𝑐²+2𝑏𝑐cos𝐴)=12(𝑏²+𝑐²−12𝑎²) 742 If 𝐴𝐷 bisect the angle 𝐴 of a triangle 𝐴𝐵𝐶, tan𝐵𝐷𝐴=cot𝐵−𝐶2=𝑏+𝑐𝑏−𝑐tan𝐴2 743 𝐴𝐷=2𝑏𝑐𝑏+𝑐cos𝐴2 744 𝐴𝐷=𝑏𝑐sin𝐴𝑎=𝑏²sin𝐶+𝑐²sin𝐵𝑏+𝑐 745 𝐵𝐷∼𝐶𝐷=𝑏²−𝑐²𝑎=𝑎tan𝐵−tan𝐶tan𝐵+tan𝐶

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive