MT - GEOMETRY - SEMI PRELIM - I : PAPER - 3

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Clifton Kelley
6 years ago
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1 07 00 MT.. ttempt NY FIVE of the following : (i) Slope of the line (m) intercept of the line (c) 3 B slope intercept form, The equation of the line is m + c ( ) The equation of the given line is + 3 (ii) tan 3 [Given] But, tan 60 3 tan tan 60 MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model nswer Paper Ma. Marks : (iii) Equation of a line parallel to Y-ais and passing through the point (3, 4) is 3. (iv) + 90º [Given] (90 ) cosec [Given] sec cosec (90 ) [ sec cosec (90 )] sec a cosec sec (v) + 3 ( 5) Comparing with the equation of a line in slope point form, m ( ) m Slope of the line + 3 ( 5) is

2 / MT (vi) + 90º [Given] tan 3 4 [Given] cot tan [ cot tan (90 )] cot Solve NY FOUR of the following : (i) (naltical figure) O 3. cm R O 3. cm R mark for circle mark for tangent (ii) The terminal arm passes through P, 3 and 3 r r units

3 3 / MT Let the angle be sin cos tan r r 3 cosec sec 3 3 cot r r 3 3 (iii) Let (3, 4) (, ) and m 5 The equation of the line passing through and having slope 5 b slope point form is, m ( ) 4 5 ( 3) The equation of the line passing through the points (3, 4) and having slope 5 is 5 0. (iv) L (naltical figure) L O 3.6 cm M 3.6 cm N N M mark for drawing circle mark for drawing tangent

4 4 / MT (v) L.H.S. sec + cosec sec, cos ec cos sin cos sin sin + cos cos. sin cos. sin [ sin + cos ] sec. cosec R.H.S. sec + cosec sec. cosec (vi) Let, (, 3) (, ) B (4, 7) (, ) The line passes through points and B The equation of the line b two point form is ( ) ( 3) [Dividing throughout b ] is the equation of the line passing through (, 3) and (4,7).3. Solve NY THREE of the following : (i) (naltical figure) C 3.5 cm 3.5 cm 7.3 cm B D

5 5 / MT C 3.5 cm M 7.3 cm B 3.5 cm D mark for circle mark for perpendicular bisector mark for tangents (ii) tan sin cos sin cos...(i) + tan sec + () sec + sec sec sec [Taking square roots] cos sec cos sin [From (i)] cosec sin cosec

6 6 / MT sin cos sec cos ec 4 sin cos sec cos ec (iii) Let, (, ) (, ) B, 3 (, ) C (0, k) ( 3, 3 ) Points, B and C are collinear Slope of line B Slope of line BC k 3 0 k 3 k 3 k + 3 k 4 The value of k is 4.

7 7 / MT (iv) 3 tan 4 3 tan tan 4tan tan 4tan tan 3tan tan tan tan 3 tan 3 0 tan 3 3 tan 0 tan 3 0 OR 3 tan 0 tan 3 3 tan But, tan 60 3 tan 3 tan tan 60 But, tan tan tan (v) Let, (, ), B ( 9, 6), C (, 4), D (6, 9) Slope of a line 6 Slope of side B 9 ( ) Slope of line B Slope of line CD ( ) 5 6 Slope of line CD 5 8 Slope of line B and slope of line CD are equal. line B line CD The line joining (, ) and ( 9, 6) is parallel to the line joining (, 4) and (6, 9).

8 8 / MT.4. Solve NY TWO of the following : (i) L.H.S. ( + tan ) + ( + cot ) + tan + tan + + cot + cot + tan + + cot + tan + cot sec + cosec + (tan + cot ) [ + tan sec, + cot cosec ] sec + cosec + sin cos cos sin sin + cos sec + cosec + cos sin sec + cosec + cos sin [ sin + cos ] sec + cosec + sec cosec (sec + cosec ) R.H.S. ( + tan ) + ( + cot ) (sec + cosec ) (ii) (naltical figure) R R 6 cm I 6.5 cm 6 cm 6.5 cm I S M 7 cm T S M 7 cm T mark for drawing triangle mark for drawing angle bisectors mark for drawing perpendicular mark for incircle

9 (iii) L.H.S. 9 / MT 3 cos sin tan sin cos 3 cos sin sin sin cos cos 3 cos sin cos sin sin cos cos 3 3 cos sin + cos sin sin cos 3 3 cos sin cos sin cos sin 3 3 cos sin cos sin (cos sin)(cos + cos. sin + sin) (cos sin) cos + sin + sin. cos + sin. cos [ sin + cos ] R.H.S. 3 cos sin + + sin. cos tan sin cos.5. Solve NY TWO of the following : (i) (naltical figure) T E 4.9 cm 0º H 6.3 cm M

10 0 / MT T E 4.9 cm 0º 6.3 cm H M 3 4 mark for drawing analtical figure mark for MT mark for constructing 7 congruent parts mark for constructing H 5 M 7 mark for constructing EH TM (ii) E Let E be the position of the cloud and let BC represent the surface of the lake. D 60º 30º Let be the point of observer 60 m and let F be the reflection of 60 m the cloud B C EC CF Let EC CF m BCD is a rectangle [B definition] B CD 60 m [Opposite sides of rectangle] EC ED + DC [E - D - C] F ED + 60 ED ( 60)m lso, DF DC + CF [D - C - F] DF (60 + ) DF ( + 60) m

11 3 In right angled DE, tan 30º ED D 60 D / MT [B definition] D 3 60 m In right angled DF, DF tan 60º [B definition] D ( 60) 3 ( 60) The height of the cloud above the lake is 0 m. (iii) (5, 4), B ( 3, ), C (, 8) seg D is the median of seg BC D is midpoint of seg BC D + +, 3 + +( 8), 8, 0, (, 5) B two point form, The equation of median D ( ) 4 ( 5) ( 5) 6 ( 4)

12 / MT [Dividing throughout b 3] The equation of median D is Slope of line C Slope of parallel lines are equal Slope of the line parallel to line C is 3 The line passes through B ( 3, ) The equation of the line parallel to line C passing through point B b the slope point form is m ( ) ( ) 3 [ ( 3)] + 3 ( + 3) The equation of the line parallel to C passing through point B is

MT - GEOMETRY - SEMI PRELIM - I : PAPER - 5

07 00 MT MT - GEOMETRY - SEMI PRELIM - I : Time : Hours Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) Slope of the line (m) 0 y intercept of the line (c) By slope intercept