圖片download連結:http://webo.com.hk/download.php?i=133684
多謝!
更新1:
不懂的數學題是所框著的題目.
更新2:
需download並解壓縮它.
更新3:
正確答案分別是2-√3 和 321.4m. 但我不知怎樣得出答案...
更新4:
wingngam第一條計錯了...
20點！2條數學題不懂
2009-05-29 1:34 am
2條有關圖形的數學題不懂，請指教。
圖片download連結:http://webo.com.hk/download.php?i=133684
多謝!
圖片download連結:http://webo.com.hk/download.php?i=133684
多謝!
回答 (5)
2009-05-30 9:56 am
✔ 最佳答案
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
2009-06-10 3:00 am
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]
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]
2009-05-30 6:57 pm
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]
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]
2009-05-29 5:47 am
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]
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]
2009-05-29 2:34 am
睇唔到圖,sorry
收錄日期: 2021-04-23 23:00:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090528000051KK01322