1. A rod with length 150cm is divided into 3 parts, a triangle with angles 30°,60°,90° is formed. Find the length of each part of rod. (express the answer in 3sig.fig.)
2. Fireworks is launched upward by ship C, and explode at point D. The angle of elevation from A to D is 30°, the angle of elevation from boat B to D is 50°, A, B and C are collinear, the distance between A and B is 90 m . Find the distance between C and D. (express the answer in 3sig.fig.)
3(a) Simplify (√a+√b)(√a-√b)
(b) Hence, rationalize 1/√2-√7
急....中二級數學
2012-08-18 7:29 am
回答 (2)
2012-08-18 10:26 am
✔ 最佳答案
圖片參考：http://i1096.photobucket.com/albums/g340/pingshek/2012-08-18.png
http://i1096.photobucket.com/albums/g340/pingshek/2012-08-18.png
參考: miraco
2012-08-18 9:37 am
HINTS:
1.
Let
A
B⊿C
angle A = 60°
angle B = 30°
angle C = 90°
note that AB+BC+CA = 150
sin A = sin 60° = BC/AB = √3/2
sin B = sin 30° = CA/AB = 1/2
sinA / sin B = BC / CA = √3
you can rewrite CA = AB/2 from sin B.
Then just solve two unknowns and two equations
2. note that BC = AC + 90
tan A = tan 30° = CD/AC
tan B = tan 50° = CD/BC = CD/ (AC+90)
solve two unknowns and two equations again.
3. (a) use identity (a+b)(a-b) = a^2 - b^2
(b) what is the question? 1/(√2-√7 ) OR (1/√2)-√7? they are different.
if the former is correct, then just multiply (√2+√7)/(√2+√7) and use the result of part (a).
1.
Let
A
B⊿C
angle A = 60°
angle B = 30°
angle C = 90°
note that AB+BC+CA = 150
sin A = sin 60° = BC/AB = √3/2
sin B = sin 30° = CA/AB = 1/2
sinA / sin B = BC / CA = √3
you can rewrite CA = AB/2 from sin B.
Then just solve two unknowns and two equations
2. note that BC = AC + 90
tan A = tan 30° = CD/AC
tan B = tan 50° = CD/BC = CD/ (AC+90)
solve two unknowns and two equations again.
3. (a) use identity (a+b)(a-b) = a^2 - b^2
(b) what is the question? 1/(√2-√7 ) OR (1/√2)-√7? they are different.
if the former is correct, then just multiply (√2+√7)/(√2+√7) and use the result of part (a).
參考: me
收錄日期: 2021-04-30 16:57:41
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