Spherical Astronomy Important Questions PDF

## The Celestial Sphere ### 127 **Ex. 3.** What is the point on the celestial sphere whose latitude longitude. R.A. and decl. are all zero. **Sol.** The point is first point of aries y or vernal equinox. **Ex. 4.** When the R.A. of the sun is (i) 6 (ii) 12h, what are its longitude and latitude...

## The Celestial Sphere ### 127 **Ex. 3.** What is the point on the celestial sphere whose latitude longitude. R.A. and decl. are all zero. **Sol.** The point is first point of aries y or vernal equinox. **Ex. 4.** When the R.A. of the sun is (i) 6 (ii) 12h, what are its longitude and latitude. **Sol.** Ref. fig. of 5.7, when the R.A. of the sun is 6" it is at point M on the ecliptic so its longitude and latitude are respectively 90° and 0°. Again when the R.A. of the sun is 12" it is at point *nie*. the point of Libra, here its longitude and latitude are respectively 180° and 0°. **Ex. 5.** If the zenith distance z of a star is less than the colatitude C, prove that $C = x + cos^{-1}(cos z sec y)$, where tan x = cot & cos H, sin y = cos & sin H, H being hour angle. [Agra 1986, 83; Meerut 81(S)] **Sol.** Let X be the star whose zenith distance ZX = z. Colatitude is given to be C. PZ = 90° − φ = complement of latitude = colatitude = C. We are given that ZX is less than PZ, therefore perpendicular XM from X on the meridian ZP will fall between P and Z. Let PM = x; .. MZ=C-x and MZ = y Also ZZPX= H and PX = 90° – δ. Now from right-angled.APMX, by Napier's rule of circular parts taking H as the middle part and x, 90° – 8 as the adjacent parts, we have sin (90° - H) = tan x tan {90° – (90° – 8)} or cos H = tan x tan 8 or tan x = cot 8 cos H. ...(1) Again from the right-angled APMX, taking y as the middle part and H, 90° – 8 as the opposite parts, we have sin y = cos (90° – H) cos {90° – (90° – δ)} or sin y = sin H cos δ. ...(2) Again from the right-angled AZMX, taking ZX (2) as the middle part and C-x, y as the the opposite parts, we have sin (90° - z) = cos (C-x) cos y or cos z = cos (C - x) cos y or cos (C-x) = cos z sec y or C = x + cos¯¹ (cos z sec y) . ...(3) Equations (1), (2) and (3) give the required result. ** (b)** Two stars (21,81) and (21,81) have the same longitude prove that! Sin (21-22) = tane (cosa.tands - Cosa, tordi) **sol-**Let X₁ and X₂ be the two stars having the same longitude (say). Also we are given that right ascension of star X₁ is &, and that of X₂ 15 22. Je, YD₁ =2, and y D₂=22 :TD, - go'td, and TD₂-90+22 Hence, <KPX₁ = 90°+2, and <KPX₂ = 90°+2₂. Again YL = X or LPKX₁ = 90°-入 KP = E(The obliquity of the ecliptic) X, D. =8, and X2D2 = 82 ∴ PX₁ = 90'-8, and P.X2 = 90°-82. Now from AKPX₁, by co-tangent formula, we have COSE cos (90+21) = Sin E cot (90-8₁)-sin(90-21) / cot (90-2₁) ⇒-cosesind₁ = sin Etandi-cosa, tanx ⇒ cos Esind₁ = - Sine tand, + cosa, dand - ① Similarly from AKPX2, we have, cos & Sind2=-sin&tand₂ + cosa, Jand-2 Now multiplying (1) by cosd₂ and eq" ② by cosα, and subtracting eq" - fran ② we gely cos(sind, 80522-cosa, Stnd₂) =-Sine (cosL2Fandy-costaner) ⇒ cose Sin (2-2) = Sine (cosa, tandy-cost,λανίδι) ⇒ Sin (21-22)=tar&(cosd, Landy-cosazdanδι). ### 150 **5.14.** Rate of Change of Zenith Distance and Azimuth [Meerut 1995, 94] Let X be the position of the star at the instant when its declination is 8 and hour angle is H. Let its azimuth be A and zenith distance be z. Now from APZX, PX = 90° - S, ZX = 2, PZ = 90° - $, ZZPX = H, and ZPZX = A. Now by applying the cosine formula in APZX, we have cos ZX = cos PX cos PZ + sin PX sin PZ cos ZPX or cos z = cos (90° – 8) cos (90° – $) + sin (90° – 8) sin (90° – $) cos H or cos z = sin 8 sin $ + cos & cos & cos H Differentiating (1) w.r.t. H, & and 8 being constants, we have - sin z dz/dH = cos o cos δ sin H or dz/dH = cos o cos δ sin H / sin z. ...(1) Again by the sine formular in APZX, we have sin z/sin (90° – 8) = sin H/sin A cos 8 sin H or sin A = sin z ...(2) ...(3) Hence from (2) and (3), we have dz/dH = sin A cos Φ ...(4) Now from (3), sin A sin z = cos 8 sin H. Differentiating above w.r.t. H, 8 being constant, we have cos A sin z dA/dH + sin A cos z dz/dH = cos & cosH. ...(5) By putting the value of dz /dH from (4) in (5), we get cos A sin z dA/dH + sin A cos z sin A cos Φ = cos & cos H or cos A sin z dA/dH = cos & cos H − sin A cos z cos Φ. ...(6) Again from APZX by the sin-cosine formula, we have sin (90° – 8) cos H = sin (90° – $) cos z − cos (90° – $) sin z cos A ...(7) cos & cos H = cos o cos z - sin sin z cos A. Therefore by putting the value of cos 8 cos H from (7) in (6), we get cos A sin z dA/dH = cos cos z- sin sin z cos A - sin A cos z cos Φ = COS COS Z (1-sin 2A) - sin sin z cos A = COS COS Z cos 2A - sin o sin z cos A or dA/dH = cot z cos & cos A – sin Φ. ...(8) If dz and dA are measured in seconds of arc and dH in seconds of times, then from (4) dz = sin A cos o dH in seconds of time =15°sin A cos o dH in seconds of arc. And from (8) dA = (cot z cos o cos A − sin $) dH in seconds of time = 15° (cot z cos o cos A − sin $) dH in seconds of arc ### **ILLUSTRATIVE EXAMPLES** **Ex 1.** A known star is seen to rise at the N.E. point. Find the latitude of the place if 8 = 30°. **Sol**. cos A = sin 8 sec $. and A = 45°, 8 = 30°. :: 1 /√2 =sec φ. .. φ = 45°. [M.U.] **Ex. 2.** Three stars S1, S2, S3 have the same R.A. and their declinations are – 6, 0, + & respectively. Show that the interval between the risings of S₁ and S2 is equal to the interval between the risings of S2 and S3. **Sol**. Let h1, h2 and h3 be hour angles at their risings, then we have cos h₁ = + tan & tan δ ; cos h₂ = 0 .. cos h3 = - tan & tan δ h₁ = 180° - h3 and h2 = 90°. h₁+h3=2h2 i.e., h₁ – h2 = h2 - h3. If the R.A. of the stars be a and t1, t2 and 13 be the sideral times of their risings. Then (t − a) - (t2 − x) = (t2 − a) – (t3 − a) . [t=a+h] **Ex. 3.** In north latitude 45°, the greatest azimuth attained by a circumpolar star is 45°. Show that the stars declination is 60°. **Sol**. The greatest azimuth is sin ⁻¹ (cos 8 sec &) = 45° or cos 8 sec & = 2 = 1 2 and sin 45° = 45° (cos 8 sec φ) = 1/√2 cos 8 = , i.e. 8 = 60", ### 152 **Ex. 4.** If a star of declination & remains h hours east of the meridian of a place and takes h' hours to come due west of it, show that cos 15h cos 15h + tan² 8 = 0 and give the conditions that this may happen. [M.U.] **Sol.** h hours = 15h° and h' hours = 15h". A star stays east of the meridian for h hours. cos 15h = - tan φ˚ tan 8° When the star is due west, cos 15°h' = tan 8º cot φ on eleminating & between the two equations, we have cos 15h cos 15h' = - tan - 2 8. ...(i) ...(ii) Here 15h must be an obtuse angle. **Ex. 5.** If y is the angle which a star makes at rising with the horizon, prove that $cos y = sin o sec 8$. **Sol.** Let X be the star at rising. XX' is diurnal path of the star. y is the angle which the star makes with the horizon while rising, i.e. considering the figure of eastern horizon LEXX' = y Also ∠PXX' is 90°, since P is the pole of the equator TR and XX' a small circle parallel to TR. Therefor, LPXN = 180° – ∠EXX' – ∠PXX' = 180° – 4 – 90° = 90° – Ψ. Also,PN = 0, PX = 90° – δ and LPNX=90°. Now from APNX, by the sine formular, we have sin (90° – 8)/sin 90° = sin (90° – ψ)/sin φ or sin (90° – ψ) = sin (90° – 8) sin φ or cos y = sino sec 8. **Ex6.** If H is the hour angle of a star at rising, show that $tan^{2} H = cos( & - δ) / cos( & + δ)$. [Meerut 1994, 91, 87, 83, 82 (P), Agra 86; Rohilkhand 79] **Sol.** We have already proved in 5.12 that when a star is rising, cos H=- tan o tan δ Now by applying componendo and dividendo, we have (equation 1 of 5.12) **5.15 Twilight** [Meerut 1996, 93, 91, 90, 85, 84, 81(S); Rohilkhand 78; Agra 84] We must have observed that it does not become dark all of a sudden when the sun sets; it is due to the sun's light getting reflected and scattered by suspended particles of dust and Water Vapour in the atmosphere of the earth. This light is called twilight. There is a similar twilight in the morning also. Evening twilight happens till the sun's center goes down 18° below the horizon, i.e. till the zenith distance does not become 108°. Similarly morning twilight begins when the sun is 18° below the horizon. Thus the interval between the sunset and the time when the sun's zenith distance has become 108° is called the *duration of evening twilight*. Similarly the interval from the instant when the sun's zenith distance is 108° to the instant when it rises is called the *duration of morning twilight*. Let S be the position of the sun when setting and S' be the position of the sun when it is 18° below the horizon. Duration of evening twilight is the time required by the sun to travel from S to S' and it is measured by the angle SPS' in hours. Let ### 184 ZZPS=H AZPS. ZS-90°. PS=90°-4. PS=PS=90°-8 ZZPS = H. cos 90° = cos (90° - 6) cos (90° – 8) + sin (90°-6) sin (90° - 8) cos H. or Again in 0 = sin sin 8 + cos & cos & cos H. AZPS, = ZS = 108*, ZZPS = H'. cos 108° = cos (90° – 6) cos (90° – δ) + sin (90° - $) sin (90° – 8) cos H cos 108* = sin o sin 8 + cos o cos & cos H'. From (1) and (2) H and H' are calculated. Thus the duration of twilight is *H-H/15* hours. By symmetry the duration of morining twilight is same. ### **Condition for twilight lasting all night.** [Agra 1984, 83(S); Meerut 1993, 85, 82; Rohilkhand 79] The twilight will last all night if the sun does not go beyond 18° below the horizon, i.e., NM < 18°, M being the point of intersection of SS' and meridian. Now PM = 90° − 8 and PN = $. Thus the condition is NM = PM – PN = 90° – 8 – φ. or 90° − 8 − ¢ < 18°, or 72° − δ < φ φ >. 72° − δ. Again we know that the declination 8 of the sun varies from 0 to the maximum value about 231/2°. Hence at places having latitude $ about or greater than 481/2°, the twilight will last all night. ### S If evening twilight ends when the Sun's centre is 18" below the horizon show that at the equator the duration of evening twilight is given in hours by B. Sint (Sin 18. Spcs). Sol't Let S be position of the sun at selling. Thus zenith distance of the sun is 90° from figure(2) os sun is on the horizon. Now from Spherical triangle AZPS, COS 90° = cos(90°-6).(os (90°-8) + Sin (90-6) Sin (30-8) cost. [Cosa = cos b.roset Sand Sine (OSA) or by cosine formula] ⇒0 = sin. Sind + cos $. Coss COSH (1) Dividing cospicos & both side we gel, ⇒0 = tong. Joan & + COSH • COSH = -tan.Sand Since at the equator f = 0 .. from en- hour angle af selling is obtained by putting f20. te COSH=0 or H=90'. Again fel s' be the position of the sun at the end of the evening twilight, Say of duration T. Now in Spherical Astangle APZS" Ify] ZS' = 90°+ 18 = 108" ∠705° =30+T ∴ Frem AP7s', we have Cos108° = cos(90-$)cos(30°-8) +sin(90-6)sin(90-8)cos(30°4T) [hy cosine formula] ⇒cos (90° +18°)=sinf-Sing - coso coss Sint > -Sin 18° = sing.sing - Cospicoss SinT. _at the equator by pulling of = 0; we get ⇒-Sin 18° =-COS&SinT.. ⇒ Sint = Sin 18° Sec8. T = Sin-1 (Sin 18'Secs) radrans T = 1x180 Sin-1 (Sin18'secs) hours 15 T = 12 sint (Sin18'secs) hrs. ### The Celestiat Sphere 187 **Ex. 4.** Show that duration of evening twilight at a place of latitude & during eunoxes is given in hours by *12/π sin (sin 18° sec $)*. [Meerut 1994, 92] **Sol.** In Ex. 3, at equator we put $ = 0; here because of equinoxes, put 8 = 0; thus we get - sin 18° = - cos o sin x sin x = sin 18° seco x = sin⁻¹ (sin 18° sec $) radians = 180 sin⁻¹ (sin 18° sec $) degrees 1 180 x 15 sin⁻¹ (sin 18° sec $) hours π 12 = sin⁻¹ (sin 18° sec $) hours π **Exes** Prove that the duration of twilight at the equator during the equinoxes is 1 hour 12 minutes. **Sol.** In this case $ = 0 at the equator and 8 = 0 at equinoxes. Thus proceeding as in Ex. 3, we get sin 18° = - sinx or x = 18° x = hours = 1 hour 12 minutes. **Ex. 6.** If at an equinox, the duration of twilight is h hours, show that the latitude of the place is *1 /cos (sin 18° cosec (πh / 12))* [Meerut 1984(P)] Sol. Let H be the hour angle at the begining of twilight (evening), so that Z.D. = 90° and the duration of twilight be T so that Z.D. = 90° + 18°, than hour angle is H + T. By cosine formula on triangles PZS and PZS (Ref. fig. Ex.3), we have cos 90° = sin $ sin 8 + cos & cos 8 cos H. sin 18° = sin & sin 8 + cos & cos & cos (H + T) . ...(1) ...(2) We know that at an equinox 8 = 0; . . sin 8 = 0 and cos 8 = 1 and hence from (1), cos & cos H = 0, cos $ ≠ 0; .. from (2), cos H = 0 or H = 90° *. in that case, $ = 90° or sin 18° = cos & cos (90° + T) = − cos o sin T. * = cos⁻¹ (sin 18° cosec T) = cos⁻¹ (sin 18° * cosec πh / 12) •∴T = Puration of twlight = 15h degree =Tx15h = π/12 ### 6.3. Refraction of a Star near the zenith. The atmosphere is considered to be made up of a large number of thin spherical layers concentric with the earth's surface which is regarded a spherical. But if we consider a star which is nearly overhead then in this case we can ignore μ sin z = un sin Z sin 2 = µη sin 2n (since µ = 1, the refractive index of vacuum) µη sin Ζη = µη - 1 sin Zn-1 μη - 1 sin 2η - 1 = µn-2 sin 2n-2 ... ... ... ... From the above equations, we get μ₁ sin z₁ = μ₁ sin ζ. sin z = µo sin ζ or sin (ζ + R) = µo sin ζ (since z = ζ + R) or Now since R is small angle, thereofre we can write cos R = 1 and sin R = R; sin ζ cos R + cos ( sin R = µo sin ζ. then above equation reduces to sin ζ + R cos ζ = μo sin ζ or R cos ζ = (μο – 1) sin ζ or R = (μο − 1) sin ζ. Putting (µo – 1) = k, this equation reduces to R = k tan ζ *Refraction = k tan (App. Z.D)* . Note. k is called the constant of mean refraction, and its value is found to be approximately 0.00029 By means of astronomical observations its value is found to be approximately 58″.2. ### 6.4. Cassini's Hypothesis. [Meerut 1991.88.86.83 82 81(S) 8O/SU ### 2021 6.9. Simpson's Hypothesis. If the relation between randu is the run+1 = constant, prove that [Meerut 1993, 89 84(S); Agra 83] 1 R = -- sin 1 sin ζ n μο where is the apparent zenith distance, a the radius of the earth and to the refractive index for lowest layer. or From the relation run +1 = constant, on differenting logari-thmically, we get dr/r + (n + 1) dμ/μ = 0. ...(1) Also, we have ur sin $ = constant. On differentiating logarithmically above relation, we have dr/r + 1/μ cot φ dφ = = 0. ...(2) Subtracting (2) from (1), we have ndμ/μ = cot φ dφ dµ 1 tan =αμ. μ Also from 6.8 equation (3), we have n ...(3) ...(4) Hence from (3) and (4), we have dR/dμ = 1/μ tan φ. ...(5) Integrating above, we get 1 R = ∫ 1/μ tan φ dφ n = 1/n * {ζ (ζ - z)}. ...(6) where z if the true zenith distance in vacuum, i.e. at the top of the atmosphere. If r be the radius of the atmosphere at the top where μ = 1, then µr sin $ = 1 sin z = µņa sin ζ and i.e. r = αμ +1 μ" +1 = r. 1+1 = αμ +1 = Now putting the value or from (7) in (6), we get n+1 sin z = αμό sin z = μņa sin ζ sin ζ μό 1 sin ζ z = sin n Putting the value of z from (8) in (5), we get R=1 n - sin This is called Simpson's formula. VVE 610 Brodloula F 1 sin ζ μο V ...(7) ...(8) ...(9) ### Refraction 207 √(μο) sin ζ = sin (ζ + R). 2 Ex. 2. If we express the refraction as K'tanz, where 2 is the true zenith distance instead of k tan & where is the apparent zenith distance, show that if k, k' are expressed in seconds of arc, then k' = k (1-k sec² & sin 1"). Sol. We know that z = ζ + R, where z and are true and apparent zenith distances respectively and R the refraction. Again R = k tan ζ = k'tan z (given) . k tan ζ = k'tan (ζ + R) = k' (tan + tan R) / (1 - tan & tan R) tan + R sin 1" = k' 1 – R sin 1" tan ζ since R is small, we have taken tan R = R = R sin 1" where R is measured in seconds of are or = k tan ζ + ktan Ϛ sin 1” k=k ог k' = k 2 1 − k tan sin 1'' 1 + k sin 1" 1-k tan2 2 ζ sin 1” " 1 − k tan², sin 1” 1 + k sin 1" = k (1 − k tan 2 2 (since R = k tan ζ) sin1") (1 + k sin 1")¯¹ = k (1 − k tan sin 1") (1 - k sin 1") 2 = k (1 − k tan² 2 (neglecting k² and higher powers) tan ( sin 1" – k sin 1") approximately " = k { 1 − k sin 1” (1 2 = k (1 − k sin 1” sec - + tan ² ζ)} 2 c² ζ). 2 ω = n! -1 ζ - ζ - (ω² – ω) tan 2 tan³ ζ 000-000 2 2 ω ω² tan ζ 6.11. Effects of Refraction. 2n 15-2 tan³ ζ. い (neglecting powers higher than w²) Effects of Refraction on Sunrise and Sunset. [Meerut 1986, 82; Agra 83; Rohilkhand 78] When the sun is on the horizon at rising or setting, then the zenith distance is 90° and horizontal P 90°- Z Η+ΔΗ refraction is 35'. We know that if there be no refraction, then the hour angle H of the sun (centre of the sun), when rising is given by ...(1) cos H = - tan o tan 8, where is the latitude of the place and 8 is the declination of the sun. On account of refraction the sun will appear to rise when its true zenith distance is 90° 35' and let H + AH be the hour angle when the sun's centre appears to be on the horizon. Now from APZX by the cosine formula, or. 90°-δ X Fig. cos (90+35') = sin o sin 6 + cos o cos & cos (H + AH) - sin 35′ = sin $ sin 6 + cos & cos 8 (cos H.1 − sin Η ΔΗ) 90°35′ * f(x+h) = f(x) + hf'(x)+21 f" (x) + ... X f" (x) = (1 - x2,3/2 etc Here f(x) = sin¯¹ x, f'(x) = * 1 √(1-x²) When a celestial body is on horizon, then the refraction is called horizontal refraction and its value is 35'. -COS& cos & sin II. AII In minutes of time Remeber [cos // tantan & from (1)] 35 sin 1' = 15 sin 1' cos & cos & sin II AL ΔΙ 35 15 sec sec & cosec // in miniutes. A minutes of time = 15 minutes of arc -15n sin 1' in circular measure. 6.12. Effect of Refraction in Right Ascension and Declination. [Agra 1985(S), 83; Meerut 1996, 92, 87, 85, 81] Let X be the ture position of the star and X' be its position on ZX on account of the refraction XX' = k tan ζ. where is the apparent zenith distance ZX', Let the hour angle and declination of X be H and 8 and that of X' be H' and 8'. Join PX and PX' and let them meet the equator in A and B respectively, when produced, Through X' draw perpendicular X'D on PX. Now regarding the small triangle XX' D as a plane triangle, we have : X'D = XX' sin n = k tan ζ sin η, XD = XX' cos η = k tan ζ cos η. The hour angle of X' is diminished by Z XPX'. Refraction in right ascension; or Δα = α' - α = γ Β - γΑ = ΑΒ = X'D sec X'B* Δα = k tan ζ sin η sec δ' [from (1) and X 'B = δ'] Refraction in declination: or Δδ = δ' - δ =BX' - AX = AD-AX= DX. Δδ = k tan ( cos η [from (2)] Again sidereal time ΖΧΡΧ' = – ΔΗ. N P 90°- Z X'(8') 90%8D% (2,5) η W T T A 8 ...(1) ...(2) = H.A. + R.A. α + Η = α΄' + H' or α- α = Η - Η′ = − (H′ − H) or Η' – H = − (α΄ – α) ог ΔΗ = − Δα = − k tan ζ sin η sec δ΄. Fig. X'D is small circle arc through X' parallel to great circle arc AB and therefore as we know X'D = AB cos X'B. .:. XX' = k tan z. In the above relation, 8' may be replaced by 8, since refraction is very small. ∆H = − k tan & sin η sec δ. Now, we observe that in the values of Δα, Δδ, ΔΗ there occurs one unknown quantity n which is found by the fact that XX' being small, we can take η without any loss of accuracy as the parallactic angle PXZ. or or Now from APXZ by the cosine formula, we have cos (90° - $) = cos ZX'cos PX' + sin ZX' sin PX' cos PX 'Z sin ¢ = cos ζ cos (90° – δ') + sin ζ sin (90° – δ') cos η sin ¢ = cos ζ sin δ' + sin ζ cos δ' cos η ### 6.13. Effect of refraction on the positions of two neighbouring stars. X₁ and X2 are two neighbouring stars, the true angular distance between them being D (in seconds of arc). If Z is the zenith, ZZX1X2 = ψ and the true zenith distance of X₁is z. Prove that the observed angular distance is *D− k D (1 + cos y tan² z)* in which k is mea Medit 1988, 67; 85%Rohilkhand 79) In order to prove the above result, first of all we deduce the following resuit ### To prove that an account of refraction the circular dise of the Sun appears to be an ellipse. Books-fet s be the centre of the sun whose radius is a. fel 25=z, and take a point P on its - limb. Due to refraction Let P be displaced to p'.. Now from P and p" drow ird pa Since Po is small the zenith distance of P and Q are same, Thu's 'P'O" IS the displaced position of Po due to refraction. Now taks S. as origin and Zsas axis of x and a line through S and jod to ZS as y=alls. Take (x,y) as the co-ordinates of p' referred do this set of arces, bel a9 Consider LPSZ = 0. Now, x = so' = 50+00 ['200' 15 = acoso + ktan 29 ∴001=klan Z0. = acoso + ktan (SZ-SQ) = acoso+ktan (2-91050/ = acoso + K[tanz-acosOser2z7 By Taylorls theoren/ x-latanz = acoso(1-K Secz) - ① y = P'o'-PO-KPQ=(1-k)PQ =(1-k)@ino[:pa = asino ] - Eleminating o from egh and from egn- coso = Tktanz③ and fromien 2 9(1-ksechz) Seno = adj. - 1 9(1-k), Squaring and adding eqn③ and (4): we get g2 tanz) 2 1 02(1-1k)2 a²(1-ksecz)2 on of which is an ellipse having centre ### Ex. 6. If the horizontal refraction be 35', show that the formula for the hour angle H of the sun's centre at rising or setting when its declination is & is cos² (11/2) = sec 8 sec & 8 cos 1945° 45" +17.5'- 2 2 8 X 45°- 17.5- 22 Sol. We know that ...(1) cos z = sin o sin 8 + cos & cos & cos H. Since refraction is 35' therefore the true Z.D. of the centre of the sun at rising or setting will be 90° 35'. Therefore from (1) or cos (90°35')