題目係 :
已知測量桿BC高1米 , 大廈DE高90米 . 若測量點A與大廈相距270米 , 問測量桿與大廈的距離是多少?
中三數一題 , 要有詳盡回答
2007-12-17 7:48 am
回答 (8)
2007-12-17 8:03 am
✔ 最佳答案
先証 △ABC ~ △ADE∠CBA = ∠ EDA (已知)
∠CAB = ∠ EAD (同位角)
∴△ABC ~ △ADE (A.A.)
ED:CB=DA:BA (相似三角的對應邊)
90 /1=270/BA
即 BA=3
2007-12-17 00:04:31 補充：
先証 △ABC ~ △ADE∠CBA = ∠ EDA (已知)∠CAB = ∠ EAD (同位角)∴△ABC ~ △ADE (A.A.)ED:CB=DA:BA (相似三角的對應邊)90 /1=270/BA即 BA=3DA-BA = 270-3 = 267
參考: 自己做既 - -
2007-12-19 4:18 am
it is not good. i have a bettemethod
2007-12-17 7:24 pm
ABC相似ADE(AAA)
DE/BC = 90:1
DA/BA = 270:x
x=270/90
=3
DA - BA = DB
270 - 3 = 267
DB=267
測量桿與大廈的距離是多少267m
DE/BC = 90:1
DA/BA = 270:x
x=270/90
=3
DA - BA = DB
270 - 3 = 267
DB=267
測量桿與大廈的距離是多少267m
參考: me
2007-12-17 6:01 pm
tan angleEAD = tan angleCAB
so, ED/AD = CB/AB
90/270 = 1/AB
hence AB = 3m
so DB = AD - AB = 270-3 = 267m
呢個應該係最簡單既方法
so, ED/AD = CB/AB
90/270 = 1/AB
hence AB = 3m
so DB = AD - AB = 270-3 = 267m
呢個應該係最簡單既方法
參考: 自己計
2007-12-17 8:47 am
tanㄥCAB=ED/DA
=90/270
ㄥCAB=18.4度(corr.to 3.s.f)
tan18.4度=CB/BA
BA=1/tan18.4度
BA=3.00m(corr.3 s.f.)
Because,DA-BA=DB
So,DB=270-3=267m
=90/270
ㄥCAB=18.4度(corr.to 3.s.f)
tan18.4度=CB/BA
BA=1/tan18.4度
BA=3.00m(corr.3 s.f.)
Because,DA-BA=DB
So,DB=270-3=267m
2007-12-17 7:59 am
測量桿與大廈的距離是267米
2007-12-17 7:55 am
AB/BC = AD/DE
AB/1 = 270/90
AB= 3m
then BD=270-3=267m
AB/1 = 270/90
AB= 3m
then BD=270-3=267m
收錄日期: 2021-04-20 00:46:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071216000051KK05092