300. If RO be produced to cut the circle again in L, then TP° = TR.TL [Euc. III. 36] =TR (RL + TR) = 2TR. RO + T'R”. But TR will in general be much less than a thousandth part of RO, and therefore TR will be much less than a thousandth part of 2TR. RO. Hence, the formula TP2 = 2TR. RO, i.e. TP' = twice the earth’s radius x vertical height, will give the value of TP correct to at least three significant figures. Example. Three times the height in feet of the place of observation above the sea is equal to twice the square of the distance of the horizon in miles. Here, TP2=RL ~ 7914 miles. Let f be the number of feet in RL, then the number of miles in RL is ; let x be the number of miles in TP, then 5280 n EXAMPLES. LXXV. 211 (1) Show that the limit of Ra sin (i.e. the area of a poly 2 gon of n sides inscribed in a circle of radius R), when n=oo is aR?. n (3) Given that 7=3•141592653589793...prove that the circular measure of 10" is •00004848136811... (4) Prove that 2 sin (72° + A) – 2 sin (72o – A)=(1/5 – 1) sin A, and that 2 sin (36° + A) – 2 sin (36o – A)=(1/5+1) sin A. (5) If a mast of a ship be 150 feet high, show that the greatest distance seen from its top is 15 miles nearly. (6) Prove that if the dip of the horizon at the top of a mountain is 1° 26'[=tan-1 •025), the mountain is about 6530 feet high. NOTE. The definitions given in Arts. 75, 78 of the Trigonometrical Ratios are now used exclusively. The names tangent, secant, sine, were given originally to quantities defined as follows. Let ROP be any angle. With centre 0 and any radius describe the arc RP. Draw PM perpendicular to OR and PT perpendicular to OP. (See Figure on previous page.) Then PR is called an arc, PT is the tangent of the arc PR, OT is the secant of the arc PR, MP is the sine of the arc PR. The name sine is derived from the word sinus. For, in the figure, PMP' is the string of the “bow” (arcus), and the string of a bow when in use is pulled to the archer's breast. The co-tangent, co-secant and co-sine are respectively the tangent, secant and sine of the complement of the arc or of the angle. The sine, tangent, etc. of the angle are the same as the measures of the sine, tangent, etc. of the arc, when the radius of the circle is the unit of length. * * MISCELLANEOUS EXAMPLES. LXXVI. (1) If 2 cos 0 - cos 20=a and 2 sin 0 - sin 20=b, prove that (aż + b2 – 3)2=12 – 8a. (2) If h cos 0 + k sin 0=1 and I cos 0+ m sin 0=1, prove that (l-h) + (m - k)?=(th-mk) (3) The diagonals of a rhombus are 2a and 2b, prove that a? ~ 62 the cosines of its angles are + a+ 62 (4) One of the values of [2++ {2+7 (2+...+12+2 cos A)}] is 2 cos an (5) If 8x=loge 3, prove that the angle whose tangent is TT (6) If the number of degrees (A) in an angle is the same as the number of radians in another angle, and if the tangents of these angles are equal, prove that A is some multiple of 1807 180 - 7 (7) If a, b, y are in A.P., then sin a + sin y=2 sin ß.cos (B-a). (8) Two parallel chords of a circle, lying on the same side of the centre, subtend respectively 72o and 144o at the centre. Prove that the distance between the chords is half the radius of the circle. (9) Prove that 4 sin (0-a). sin (m8 --a). cos (0 – mo) =l+cos (28 – 2m0) - cos (20 – 2a) - cos (2m0 - 2a). (10) Express 24 +y4+* - 2y2z2 – 2z2x2 – 2x2y2 in a form fitted for logarithmic calculation. . (11) Solve the equations cos (8 + x) m sin ß (iv) tan mo=cot no. A - B (12) In any triangle tan =tan (0 2 where $=tan-1%. 62 +62 – 2bc cos (60° + A)=c2 + a2 – 2cà cos (60° +B). Hence prove that if equilateral triangles are described on the sides of the triangle ABC, the centres of circles inscribed in them will be the angular points of an equilateral triangle. (14) A round tower stands on an island in a lake. A, B are two points on the land such that AB is a feet and points directly to the middle of the tower. At A and B the base of the tower subtends the angles 2a and 2B respectively. Prove that the a sin a.sin ß. diameter of the tower is 2 sin ß - sin a (15) At P the top of a tower of height h, the angles of depression of two objects on the horizontal plane on which the tower stands are 10-a and 1 ata. Prove that the angle APB=2a, and that AB=2h tan 2a. (16) On the side of a hill there are two places B, C inaccessible to each other, but known to be at the same distance (a) from a certain station A also on the hill. At the lower place C the horizontal angle 8 between A and B is observed as well as the altitudes d, u of A and B. Prove that the distance between A (17) In the ambiguous case, given B, c, b, if az, a, are the two values of a, prove the following statements : (i) , +az=2c cos B. (ii) a,a,=c-64. (iv) The distance between the centres of the circumscribing circles of the two triangles is ay ~ az 2 sin Bo (v) The circumscribing circles of the two triangles are equal. (vi) If B=45°, the angle between the two positions of l is 2αα, COS -1 (18) Prove that if 11, 12, 13 are the centres of the escribed circles, then 12A1, is a straight line, and 1,4, 1,B, 13C are perpendicular to 1,13, 1311, 11, respectively. Prove also that (i) 1,12=a cosec 4. (ii) The angle 1,113= - 4 HS (iii) The area of 1,1213 sin A A B C. 2 2 sa + + (19) If D, E, F are the feet of the perpendiculars from A, B, C on the opposite sides, the triangle DEF is called the pedal triangle of the triangle ABC*; prove that (i) EF= '=a cos A. (ii) The angle EDF=T-2A, 1 1 1 1 AD2 bc (iv) AD BET CF BE.CF-a2 (v) The radius of the circle circumscribing A EF--Rcos A. (vi) The radius of the circle circumscribing DEF=R. * The student should notice that ABC is the pedal triangle of 1,7,12 in Question (18). |