Question

Monster Circle Problem Name: AD is tangent to OH at point C FR is tangent to OH at point G CJ is a diameter m22 -89 CB = 23° BN = 62° NM = 64° 1 - 18° CE = 63 EG = 57° K

hello I need the relationship of all the angles please, Thanks.

trigonometric

Answer 1

solution) From given conditions we have HCI AD & FX LHG .. L20+ (21 = 90°... 227 = 90° --- (Am) L24+ [25+ 226 = 90 arc Bc sublends angle 23° at centie, (given) .. U10 = 23°. (Am) Corc BN subtends angle 62 dx-centres (given) *= 62°. (Am) from dete, we have given) LQ = 899.--Am are nr subtendo angle bu a-centre (given) As BM is a straight line, this L2+ F L3= 180 = L3=180-89, I min BM & NHintersect at =) L3=91. An). thes La=L7 & L3= L8 (symmetrical angles » L7= 89° . (A) - L8=90-tam) From property of briangles we have 28+La+ Lil = 180 Am). =) L9 = 180-91-62=27. arc cE sublends angle 63 at- Centre Cainen). arc Ea sablende ongle st al- centre Cainen) this LCHG= 63 +57° - 1200 In A CHG, CH=HG Cracius) thus L.20= 226 coupe also LCHG+ L 20 +226 = 180° >> 2 420 = Go - L202 30º du) (26=30 Am 919

Now Criver CJ is diameter, :: Liat LCHG = 180° => 212= 180 - IBO° = GO...ffin) As CG is a chord to tangent- i LDCA = 2 LCHG (from tengene-chord properly) = 121= x120 = 60° - L21 = 60°. --Etu) Also L21+ L20 = 90° L20 = 30°: Similarly for LEGF= 1 LCHG. :) 224+ 125= 60.- Similarly for LEGF= 2 LENG - L24 = 1x S7 - Lau=28.5.An) in 225=60-28-5 - L25=315 An).

By the property of angle formed by 2 intersecting chords we have, L 19= 1 (MLCE + M LLI+L12) 419= L19 = 70.5" (other) s(63+ 18 + 60 ) = 70.5 218= 180 - L19 (ne on straight line) = 180 - 70.5=7 U8 = 109.5. ..Am). Similarly, L14 = 2 (m LCE+ M L LJ). =£(63+ 18) = 40.5°-om) Also, L13+ Li4= 1800 (ine on a straight line) => L13= 139.5"..(a) from propony of triangles we have, [2+* 212+ L17 = 213. - L17= 13- L12 A » L17= 79.5. .. sm). from property of quadrilaterals, [internal angles= 360° > (_21+220) + L22 + LC1G + (180 - 217) = 360° - £22 = 49.5. An).

By property of triangles we have, L23 +225= 219 » La3 = 70.5"-315 L23= 39'. - (Au) angle subtended by arc mL (minor) is obtained ky, in LML = 360 - (110 + Lll + LCHG+L12 M LJ tm LNM) -) MLM2 = 360 - (L10 + Lil +120 + L12+ L + NM) = 95- G4-18 = 95-82 = 13° By peoperty of angle subtended by chords, 24 = 1 (MLBNT ML ML) = a (G2+13) = 37.5 -> L4 = 37.5 -18h) . L4+25= 180 (ne on scraight line) => 15=142.5° (m) L6 = 25 (symmetrically opposite angles) s 26= 142-50. An). By propony of triangles we here, Lit L3 + 24 = 180° ) L1 = 180 - (37.5+91) => L1 = 57.5, By propering of quadrilateral E internal angles= 360°. we obtain, L16 = 63.5 . 25= 74. (an) cops