![]() ![]() Number of days of absent Number of students Solution : C.I. Find the mean number of days, a student was absent. For the month of February, a class teacher of Class IX has the following absentee record for 45 students. = sec q cot (90 q) = sec q = sec q (tan q) = sec q tan q = R.H.S. Solution : To prove : sec q cot (90 q) = cos (90 q) + cos q. If sec A =, find the value of tana cosa + +sina tana Solution : Given, sec A = In ABC B C A AC = AB + BC = ( ) = BC 4 = + BC BC = 4 BC = BC = So, tan A = cos A = sin A = tan A sin A = + cos A tan A Mathematics 06 Term I 45 = + = + = Prove that : sec q cot (90 q) = cos (90 q) + cos q. ![]() (ii) = Now in RQP and PST = So by AA similarity RQP ~ PST ST QP = PS RQ.(i).(ii) 4 = 5 ST.RQ = PS.PQ Hence Proved.ΔΆ 7. ![]() Solution : In D RPQ = 80 + = 90 TP PQ TPQ = 90 + =90 = 90 From eq. In the given figure, RQ and TP are perpendicular to PQ, also TS PR prove that ST.RQ = PS.PQ. (iv) AC + DE = AB +BC + BD + BE = AB + BE + BC + BD AC +DE = AE + CD Hence Proved. To prove : AC + DE = AE + CD In ABC by using Pythagoras theorem, AC = AB + BC.(i) In ABE by using Pythagoras theorem AE = AB + BE.(ii) In BCD by Pythagoras theorem CD = BD +BC.(iii) In DBE by Pythagoras theorem DE = DB + BE.(iv) Adding eq. Solution : In ABC, B = 60 and D, E are point of AB, BC respectively. D and E are any point on AB and BC respectively. ABC is a right angled triangle in which B = 90. (ii) u 4v = u + v = 9 7v = 7 v = Putting the value of v in eq. ![]() Solve the following pair of equations by reducing them to a pair of linear equations : 4 - = x y + x y = 9 4 Solution : Given, x y = Let + x y = 9 x = u, = v y So, u 4v =.(i) u + v = 9.(ii) On solving eq. 1 450 Mathematics 06 Term I So fourth zero x = 0 x = Hence four zeroes will be, 0. ![]()
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