9. Shew that the length of the straight line drawn to bisect the angle A of a triangle and terminated by the opposite side is 10. 2bc cos A b+c The angle C of a triangle is a right angle. Shew that the radius of the inscribed circle is equal to } {a+b−√√(a2+b2)}. 11. Shew that the radius of a circle which passes through the vertex A of a triangle, and touches the side BC at its middle point is 2(b2+c3)-a2 12. P, Q, and R are points in the sides BC, CA, and AB respectively of a triangle, such that shew that PQ2+QR3 +RP2 = (a2+b2+c2) (1 − 3x+3x2). XIX. Trigonometrical Ratios of two angles. 170. The object of the present Chapter is to express the Trigonometrical Ratios of the sum or difference of two angles in terms of the Trigonometrical Ratios of the angles themselves. 171. To express the sine of the sum of two angles in terms of the sines and cosines of the angles themselves. Let the angle COD be denoted by A, and the angle DOE by B; then the angle COE will be denoted by A+ B. In OE take any point P, draw PM perpendicular to OC, and PN perpendicular to OD; draw NR perpendicular to PM, and NQ perpendicular to OC. Then the angle NPR is the complement of PNR, and is therefore equal to RNO, which is equal to NOQ or A. 172. To express the cosine of the sum of two angles in terms of the sines and cosines of the angles themselves. E P R N M The same construction being made as in the preceding Article, we have 173. To express the sine of the difference of two angles in terms of the sines and cosines of the angles themselves. Let the angle COD be denoted by A, and the angle DOE by B; then the angle COE will be denoted by A – B. In OE take any point P, draw PM perpendicular to OC, and PN perpendicular to OD; draw NR perpendicular to MP produced, and NQ perpendicular to OC. Then the angle NPR is the complement of PNR, and is therefore equal to DNR, which is equal to NOQ or A. 174. To express the cosine of the difference of two angles in terms of the sines and cosines of the angles themselves. The same construction being made as in the preceding Article we have 175. To assist the student in remembering the preceding demonstrations, we may observe that the point P is taken in the straight line which bounds the compound angle we are considering; thus in demonstrating the formulæ for sin (A+B) and cos (A+B) the point P is taken in the straight line which bounds the angle A + B, and in demonstrating the formulæ for sin (4-B) and cos (A-B) the point P is taken in the straight line which bounds the angle A-B. 176. The formulæ established in Arts. 171...174 are true whatever may be the size of the angles A and B ; the student may exercise himself by going through the construction and demonstration in various cases; it will be found that the only variety which occurs in the construction consists in the circumstance that the perpendiculars in some cases fall on certain straight lines and in other cases fall on those straight lines produced. 177. We will for example demonstrate the formulæ of Arts. 173, 174, for the case in which the angle A is greater than a right angle, while B is less than a right angle, and A – B is less than a right angle. Let the angle COD be denoted by A, and the angle DOE by B; then the angle COE will be denoted by A-B. In OE take any point P, draw PM perpendicular to OC and PN perpendicular to OD; draw NR perpendicular to MP and NQ perpendicular to CO produced. Then the angle NPR is the complement of PNR, and is therefore equal to RNO, which is equal to 180°-A. Then RM+PR_NQ PR sin (4-B)= PM OP = + = + NQ ON PR PN ON OP PN OP = sin (180o — A) cos B+ cos (180o — A) sin B =sin A cos B-cos A sin B, by Art. 95. |