practice problems

Q1.


S1.
First do angle chasing and get all angles.



Now apply Sine rule in AMC.
AC/sin(150) = MC/sin(7)
sin(150) = sin(180-150) = sin(30) = 1/2
=> AC = MC/2sin(7)___[1]

In BMC:
BC/sin(M) = MC/sin(97-M)
BC = AC and sin(97-M) = sin(90 + 7 - M) = cos(7-M) = cos(M-7)
=>
AC = MC.sin(M)/cos(M-7)  ____[2]
Using [1] and [2]:
1/2sin(7) = sin(M)/cos(M-7)
=>
2sin(7)sin(M) = cos(M-7) = cosMcos7 + sinMsin7
sin7sinM = cos7cosM
tan7 = cotM = tan(90-M)
=> M = 83 = answer

Q2.


S2.


Isosceles triangle, altitude bisects the base.
x^2 = h^2 + 225
y^2 = h^2 + 36

(x+y)(x-y) = 189 = 3^3*7
189 = 189 * 1 = 63 * 3 = 21 * 9 = 7 * 27
case 1:
x+y = 189, x-y=1 => 2x = 190 => x = 95, y=94 => AD = CD = 95, perimeter = 190 + 30 = 220

case 2:
x+y = 63, x-y=3 => 2x = 66 => x = 33, y=30 => AD = CD = 33, perimeter = 66 + 30 = 96

case 3:
x+y = 21, x-y=9 => 2x = 30 => x = 15, y=6 => AD = CD = 15, invalid since AC = 30

case 4:
x+y = 27, x-y=7 => 2x = 34 => x = 17, y=10 => AD = CD = 17, perimeter = 34 + 30 = 64

final answer: 
220 + 96 + 64 = 380

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