f.4 三角學一條

The Q is solved as follows:

圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths274.jpg?t=1213522435

Let E and F be the foot of perpendicular of D and C respectively, i.e ∠DEA=90。 and ∠CFB=90。.
AE+BF+8=12
AE+BF=4
Since DE=CF, suppose there is a triangle PQR with side lengths PQ=AD, QR=BC and PR=4cm and ∠QRP=40。 and ∠QPR=60。.
Therefore, ∠PQR=180。-40。-60。=80。 (∠ sum of Δ)
Then, by sine formula,
PQ/sin40。=4/sin80。 and QR/sin60。=4/sin80。
PQ=4(sin40。)/(sin80。) and QR=4(sin60。)/(sin80。)
Thus, AD=PQ=2.6cm (correct to two sig. figs.)
BC=QR=3.5cm (correct to two sig. figs.)

P.S. This type of question is often appearing in the Maths. CE, therefore, when you are tackling these equations, you should immediately recall you memory of using sine/cosine formulae. In fact, for simplicity, you should imagine there is a triangle instead of a trapezium in order to use the above formulae.

P.S. I think this question is indeed a F.4 level question. You know how to solve this question because it is not necesary to apply sine/cosine formluae which you are going to learn in F.4.

Let a + b + 8 =12,
a+b = 4
b =4-a

atan60度=btan40
atan60 =(4-a)tan40
a(tan60+tan40)=4tan40
a =4tan40 / (tan60+tan40)
= 1.305 (如果答案要求3 sig.fig. 呢到就寫4sig.fig.)

AD=a/cos60
=1.305 / cos60
=2.61

b= 4-a = 2.695

BC = 2.695 / cos40
=3.52

所以 AD=2.61 (cm) , BC=3.52 (cm)

我用了中三的知識去計,我想這題應不是中四數=.=

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