13. In the figure, AD is the altitude from A to the side BC produced. Angle CAB=15 degrees, Angle ACD=60 degrees, BC = 100 and AD=h.
(a)Express BD and CD in terms of h.
(b)Hence, find the value of h.
Maths problems!!!!!!!!!!!!!!!!
2009-06-08 1:02 am
回答 (3)
2009-06-08 3:49 am
✔ 最佳答案
001 is wrong about ∠ABDa)
CD = h / tan 60
∠ABD = 15 + 60 = 75 (ext. ∠ of triangle ABC)
BD = h / tan (180 - ∠ABD) = h / tan 75
b)
CD - BD = BC
h / tan 60 - h / tan 75 = 100
h x tan 75 - h x tan 60 = 100 x tan 60 x tan75
h = 323.205
2009-06-08 3:37 am
i guess the 15 degree is angle ABD, right^^?
(a) tan ABD = AD / BD
so, BD = h / tan 60 = h / root(3)
and, tan ACD = AD/ CD
so, CD = h / tan 15
(b) BC = BD + CD
so, BC = h / root(3) + h / tan 15
100 = h* ( 1/root(3) + 1/tan 15 )
so, h = 100 / ( 1/root(3) + 1/tan 15 )
= ... (禁機~)
2009-06-07 20:29:22 補充：
原來係咁樣~
你(a) part tangent果度唔駛 "180 - ", 不過係小問題黎je~
同埋, (b) part 可以抽 h 出黎, 成個 ( 1/ tan - 1/ tan ) 除過去, 會快D的 :P
2009-06-07 20:30:23 補充：
002 manyu_yan 的答案確實最佳呢~
(a) tan ABD = AD / BD
so, BD = h / tan 60 = h / root(3)
and, tan ACD = AD/ CD
so, CD = h / tan 15
(b) BC = BD + CD
so, BC = h / root(3) + h / tan 15
100 = h* ( 1/root(3) + 1/tan 15 )
so, h = 100 / ( 1/root(3) + 1/tan 15 )
= ... (禁機~)
2009-06-07 20:29:22 補充：
原來係咁樣~
你(a) part tangent果度唔駛 "180 - ", 不過係小問題黎je~
同埋, (b) part 可以抽 h 出黎, 成個 ( 1/ tan - 1/ tan ) 除過去, 會快D的 :P
2009-06-07 20:30:23 補充：
002 manyu_yan 的答案確實最佳呢~
參考: 這是非常典型的三角學題目~
2009-06-08 1:32 am
Can you provide the figure?
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