f.2 maths trigonometric ratios
2010-05-02 10:35 pm
1)In the figure, the length AB of the drawbridge is 13m, and the width BC of a channel is 10m. After putting down the drawbridge, point A is right above point C. Find the turning angle θ of the drawbridge.
http://cg-love.com/image-2622_4BDD1A60.jpg
2)
The figure shows a rectangle ABCD. EAFG is a straight line. Find BF.
http://cg-love.com/image-7D91_4BDD1C26.jpg
3)
Find x and θ in the figure.
http://cg-love.com/image-B653_4BDD1BEA.jpg
4)
In the figure, ABCD is a square and AED is a straight line. Find ∠BEC.
http://cg-love.com/image-EF9A_4BDD1C4D.jpg
5)
In the figure, QRS is a straight line. Find the distance between Q and R.
http://cg-love.com/image-DCBE_4BDD1C7A.jpg
回答 (2)
2010-05-02 11:01 pm
✔ 最佳答案
1) cosㄥABC = BC/AB = 10/13
ㄥABC = 39.71514°
θ = 90 - 39.71514 = 50.28°(2dec.)
2)
AB = DC = 25
ㄥBAF = 180 - 90 - 55 = 35°
BF / AB = sin 35°
BF / 25 = sin 35°
BF = 14.34 (2dec.)
3)
x^2 + (11-6)^2 = 13^2
x = 12
12/13 = sin θ
θ = 67.38°
4)
tanㄥAEB = AB/AE = 10/4
ㄥAEB = 68.19859°
tanㄥDEC = CD/DE = 10/(10-4) = 5/3
ㄥDEC = 59.03624°
ㄥBEC = 180 - 68.19859 - 59.03624 = 52.77°(2dec.)
5)
PS / (QR + 50) = tan22°
PS = (QR+50)tan22°......(1)
PS / 50 = tan43°
PS = 50tan43°......(2)
(1) = (2) = PS :
(QR+50)tan22° = 50tan43°
QR+50 = (50tan43°)/tan22°
QR = 65.40279 m
2010-05-03 7:59 am
1)sin BAC=10/13
BAC=sin^-1(10/13)
BAC= θ
so, θ =sin^-1(10/13)=50.3
2)angle BAF=90-55=35
BA=CD=25
sin35=BF/25
BF=25sin35
=14.3
3)X=(13^2-5^2)^(1/2)
=12
cosθ=5/13
θ=67.4
4) tan ABE=10/4
BEC=tan^-1(4/10)+tan^-1(6/10)
=52.8
5)PS=50tan43
QR=((50tan43)/tan22)-50
=65.4
BAC=sin^-1(10/13)
BAC= θ
so, θ =sin^-1(10/13)=50.3
2)angle BAF=90-55=35
BA=CD=25
sin35=BF/25
BF=25sin35
=14.3
3)X=(13^2-5^2)^(1/2)
=12
cosθ=5/13
θ=67.4
4) tan ABE=10/4
BEC=tan^-1(4/10)+tan^-1(6/10)
=52.8
5)PS=50tan43
QR=((50tan43)/tan22)-50
=65.4
收錄日期: 2021-04-13 17:13:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100502000051KK00888