20點！2條數學題不懂

不是太難的....希望幫到你

1. 我計到係321.4 m
< APB
= 180 - 66 - 42
= 72

500/sin72 = PB/sin66
PB = 480.279

Let the point which between AB be C,

< CPB = 180 - 90 - 42
= 48

CB/sin 48 = 480.279/sin90
CB = 356.917

PC = (成個開方) (480.279)^2 - (356.917)^2
= 321.4 m

2009-05-30 02:13:12 補充：
2.
< BAD = 180 - 90 - 30
= 60

< ADC = 180 - 30
= 150

< CAD and < ACD = (180 - 150)/2 = 15

tan 75 = BC/AB
BC = ABtan75

tan 15 = AB/BC
tan15 = AB/ABtan75
tan15 = 1/tan75
tan15 = 0.268
tan15 = 2-開方3

1)

Let the height of the balloon be h. And Let k be the horizontal distance from A to the balloon.

then h=(k)tan 66=(500-k)tan 42

solve k=105.4738, h=236.8981

2)

tan 30=AB/BD

AB=BDtan30=DC

and, angle ACD=15

therefore, tan (angle ACD)=AB/(BD+DC)

=BD tan30/(BD+BD tan30)

= tan30/(1+tan30)

=sqrt(3)/3/[1+sqrt(3)/3]

1)

Let the height of the balloon be h. And Let k be the horizontal distance from A to the balloon.

then h=(k)tan 66=(500-k)tan 42

solve k=105.4738, h=236.8981

2)

tan 30=AB/BD

AB=BDtan30=DC

and, angle ACD=15

therefore, tan (angle ACD)=AB/(BD+DC)

=BD tan30/(BD+BD tan30)

= tan30/(1+tan30)

=sqrt(3)/3/[1+sqrt(3)/3]

1)

Let the height of the balloon be h. And Let k be the horizontal distance from A to the balloon.

then h=(k)tan 66=(500-k)tan 42

solve k=105.4738, h=236.8981

2)

tan 30=AB/BD

AB=BDtan30=DC

and, angle ACD=15

therefore, tan (angle ACD)=AB/(BD+DC)

=BD tan30/(BD+BD tan30)

= tan30/(1+tan30)

=sqrt(3)/3/[1+sqrt(3)/3]

睇唔到圖,sorry