(1)In the finure,angleABC=90degree and AB=15. it is given that 5BC=4AC.
(a) Find AC.
(b) If CD=10 and AD=5√21, prove that angleADC=90degree.
picture:http://postimg.org/image/z2d0sf161/
(2)In the figure,CB perpendicular AB,angleCAB=40 and angleDAB=20.CD=5m a-nd DB=xm.
(a) By considering triangleABC,express AB in terms of x.
(b) Find the value of x.(correct the answer to 4sig.fig.)
picture:http://postimg.org/image/n1gmtf24t/
NEED STEP,PLZ!!!
Challenging maths problem
2013-12-19 10:39 pm
回答 (3)
2013-12-20 2:29 am
✔ 最佳答案
1.a)5BC = 4AC
BC = (4/5)AC
Since angle ABC = 90⁰
AC² = 15² + BC²
AC² = 15² + (16/25)AC²
(1 - 16/25)AC² = 225
(9/25)AC² = 225
AC² = 625
AC = 25
b)
CD = 10
CD² = 100
AD = 5√21
AD² = 25×21 = 525
CD² = AD² = 100 + 525 = 625 = AC²
==> angle ADC = 90⁰
2013-12-19 18:42:21 補充：
x/AB = tan20⁰
x/tan20⁰ = AB
AB = x/0.363970234
AB = 2.747x
(5+x)/AB = tan40⁰
(5+x)/2.747x = tan40⁰
(5+x)/2.747x = 0.8391
5+x = 2.305x
5 = 2.305x - x = 1.305x
x = 5/1.305 = 3.831
checking : (5+x)/2.747x = (5+3.831)/(2.747)(3.831) = 0.8391 = tan40⁰
2013-12-20 10:59 am
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唑古僕
唑古僕
2013-12-20 4:25 am
i want ot ask whyAB=x/0.363970234 and change to 2.747x
收錄日期: 2021-04-27 20:40:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131219000051KK00047