07 00 MT MT - GEOMETRY - SEMI PRELIM - I : PAPER - Time : Hours Model Answer Paper Ma. Marks : 40 A.. Attempt ANY FIVE of the following : (i) Slope of the line (m) 4 intercept of the line (c) 3 B slope intercept form, The equation of the line is m + c ( 4) + ( 3) 4 3 The equation of the given line is 4 3 (ii) ( + tan ) ( sin ) ( + sin ) sec ( sin ) [ + tan sec ] sec cos [sin + cos cos sin ] cos cos (iii) Equation of a line parallel to X-ais and passing through the point (, 5) is 5. (iv) tan 5 + tan sec + 5 sec + 5 sec sec 6 sec 4 [Taking square roots] (v) 5 ( ) Comparing with the equation of a line in slope point form, m ( ) m 5 Slope of the line 5 ( ) is 5
/ MT PAPER - (vi) 30º [Given] sin sin ( 30) sin 30 sin ( 30) A.. Solve ANY FOUR of the following : (i) (Analtical figure) O.9 cm M O.9 cm M mark for circle mark for tangent (ii) The terminal arm passes through P ( 4, 7) 4 and 7 r ( 4)( 7) 576 49 65 r 5 units Let the angle be
3 / MT PAPER - sin cos tan r r 7 5 4 5 7 4 7 4 cosec r sec r cot 5 7 5 4 4 7 4 7 (iii) Let A (, 3) (, ), B (4, 5) (, ) The equation of line AB b two point form is, 4 3 3 5 3 3 + 3 0 + 0 (iv) The equation of the line passing through the points (, 3) and (4, 5) is + 0. L (Analtical figure) L 3.6 cm O N 3.6 cm O M N M A A mark for circle mark for tangent
4 / MT PAPER - (v) tan A + tan A tan A tan A 4 [Squaring both sides] tan A + tan A. tan A tan A 4 tan A + + tan A + tan A 4 tan A 4 tan A + tan A (vi) Let A (3, 4) (, ) and m 5 The equation of the line passing through A and having slope 5 b slope point form is, m ( ) 4 5 ( 3) 4 5 5 5 5 + 4 0 5 0 The equation of the line passing through the points (3, 4) and having slope 5 is 5 0. A.3. Solve ANY THREE of the following : (i) (Analtical figure) m M 7.5 cm l E 6 cm L
5 / MT PAPER - M 7.5 cm O E 6 cm L (ii) 3 sin 4 cos 0 3 sin 4 cos sin 4 cos 3 4 tan 3 + tan sec + 4 3 + 6 9 9 6 9 5 9 sec sec sec sec sec 5 3 mark for triangle mark for perpendicular bisectors mark for circumcircle [Taking square roots]
cot tan 4 3 6 / MT PAPER - 3 cot 4 + cot cosec + 3 4 cosec + 9 6 cosec 6 9 6 cosec cosec 5 6 cosec 5 4 [Taking square roots] (iii) Let, A (, ), B ( 9, 6), C (, 4), D (6, 9) Slope of a line Slope of line AB Slope of line AB Slope of line CD Slope of line CD 6 9 ( ) 5 9 5 8 5 8 9 4 6 ( ) 5 6 5 8
7 / MT PAPER - Slope of line AB and slope of line CD are equal. line AB line CD The line joining (, ) and ( 9, 6) is parallel to the line joining (, 4) and (6, 9). (iv) sec + tan p sin cos cos p sin cos p ( sin) p cos sin sin cos p sin cos sin sin sin sin p sin sin sin + sin sin + sin sin sin p p + p p [B Componendo-Dividendo] p p p p p sin [B Invertendo] (v) Let, A (, ) (, ) B, 3 (, ) C (0, k) ( 3, 3 ) Points A, B and C are collinear
8 / MT PAPER - Slope of line AB Slope of line BC 3 3 3 k 3 0 k 3 k 3 k + 3 k 4 The value of k is 4 A.4. Solve ANY TWO of the following : (i) L.H.S. tan sec sec tan tan(sec + ) (sec ) tan tan sec sec (sec ) tan sec sec sec (sec ) tan sec sec (sec ) tan sec(sec + ) (sec ) tan [ + tan sec ] sec tan sec tan sin cos cos cos cos sin tan sin, sec cos cos
9 / MT PAPER - sin cosec R.H.S. tan sec + + sec + tan cosec (ii) (Analtical figure) F F 5.8 cm I 5.8 cm D 65º 5.8 cm E I D 65º 5.8 cm E mark for triangle mark for angle bisectors mark for perpendicular mark for incircle (iii) sin + cos cos sin cos. cos ( sin ) ( + sin ) cos sin sin cos B theorem on equal ratios, + sin cos cos ( sin) cos sin sin cos sin cos cos cos ( sin) sin Dividing the numerator and denominator of R.H.S. b cos
0 / MT cos + sin cos cos cos + sin ( sin) cos + sin cos cos + sin sin cos cos sin cos + sin + cos sec tan PAPER - A.5. Solve ANY TWO of the following : (i) U (Analtical figure) U R R 5.8 cm 5. cm 5.8 cm 5. cm S 4.5 cm H V S S S 3 S 4.5 cm H V S4 S 5 S S S 3 S 4 mark for drawing Analtical figure mark for SHR mark for constructing 5 congruent parts mark for constructing VS 5 S HS 3 S mark for constructing UVS RHS S 5
(ii) / MT PAPER - seg AB represents the tree A AB m The tree breaks at point D seg AD is the broken part of tree which then takes the position of DC AD DC D m m DCB 60º Let DB m AD + DB AB [ A - D - B] AD + B 60º C AD ( ) m DC ( ) m In right angled DBC, sin 60º DB DC 3 3 3 3 [B definition] 3 3 3 3 DB DB 3 3 3 3 m 3 3 3 3 4 3 (3) () 3 DB DB 4 3 36 4 3 DB 4(.73) 36 DB 4.5 36 DB 5.5 m The height at which the tree is broken from the bottom b the wind is 5.5 m.
/ MT PAPER - (iii) P (, 4), Q (4, 8), R (0, 5), S (4, ) Slope of a line P (, 4) S (4, ) 8 4 Slope of line PQ 4 ( ) 4 4 Q (4, 8) R (0, 5) 4 6 Slope of line PQ 3 5 Slope of line RS 4 0 4 6 Slope of line RS 3 Slope of line PQ Slope of line RS line PQ line RS...(i) 5 8 Slope of line QR 0 4 3 6 Slope of line QR 4 Slope of line PS 4 ( ) 3 4 3 6 Slope of line PS Slope of line QR Slope of line PS line QR line PS...(ii) In PQRS, side PQ side RS [From (i)] side QR side PS [From (ii)] PQRS is a parallelogram [B definition]