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##### Math Assignment_澳洲数学assignment格式——澳洲代写数学作业（含附件模板）

Math assignment

提供Math Assignment_澳洲数学assignment格式——澳洲代写数学作业模板

Luran(Lori) Zhou

2012/2/24

Year 11 Mathematic B
Teacher: Miss Roach

This assignment is about a cyclone is bearing down on the coast of China due east of Hong Kong and due south of Shang Hai, moving directly towards the town Yi Chang. The angles and the distances about the cyclone and towns were calculated by using Trigonometric functions and application.

Q1.Choose a town between Hong Kong and Shang Hai anduse the Internet site www.wolframalpha.com to find the latitude and the longitude of all the towns.
City The latitude the longitude
Shang Hai 31° 13' 48" N 121° 28' 12" E
Hong Kong 22° 17' 2" N 114° 9" E
Yi Chang 30 °42' N 111° 16' 48" E
cyclone 22° 17' 2" N 121° 28' 12" E

Yi Chang was chosen as the town.
The websitewww.wolframalpha.com was used to find the longitude and the latitude of all the towns and cities.
The latitude and the longitude of Hong Kong:  22° 17' 2" N, 114° 9" E
The latitude and the longitude of Shang Hai: 31° 13' 48" N, 121° 28' 12"E
The latitude and the longitude of Yi Chang: 30 °42' N, 111°16' 48" E
The latitude and the longitude of the cyclone: 22° 17' 2" N,121° 28' 12" E

Q2. Use the Internet Sitehttp://www.movable-type.co.uk/scripts/latlong.html to find and record the distance and initial bearing between the towns and cyclone.

From To Distance Initial bearing
Hong Kong Shang Hai 1231km 034°33′47″
Shang Hai cyclone 994.8km 180°00′00″
Hong Kong cyclone 753.1km 088°36′38″
Yi Chang Hong Kong 978.3km 162°22′04″
Yi Chang cyclone 1379km 130°19′00″

The website http://www.movable-type.co.uk/scripts/latlong.html was used to find all the data above.

Q3. Using measurements and sine and cosine rule to find all angles from the above diagram.

The cosine rule formula was used to calculate∠A:
cosA=(b^2 〖+c〗^2-a^2)/2bc
cos∠A=(〖978.3〗^2 〖+1379.0〗^2-〖753.1〗^2)/(2×978.3×1379.0)
∠A=〖cos〗^(-1) ((〖978.3〗^2 〖+1379.0〗^2-〖753.1〗^2)/(2×978.3×1379.0))
∠A=31°51'50″

The sine rule formula was used to calculate ∠ADBand∠ABD:
a/(sin A)=b/sinB=c/sinC

753.1/(sin31°51'50″)=1379.0/(sin∠ABD)
∠ABD=sin^(-1) （(1379.0×sin⁡〖31°〖51〗^' 〖50〗^″ 〗)/753.1）
∠ABD=75°9^' 34″
According to diagram, ∠ABD is obtuse
So ∠ABD=180°-75°9^' 34″=104°〖50〗^' 26″

Verify the angles:
In △ABD, the three angles can be added into 180°.

The cosine rule formula was used to calculate∠A:
cosA=(b^2 〖+c〗^2-a^2)/2bc
cos∠C=(〖CD〗^2 〖+CB〗^2-〖BD〗^2)/(2×CD×CB)
cos∠C=(〖994.8〗^2 〖+1231〗^2-〖753.1〗^2)/(2×994.8×1231)
∠C=〖cos〗^(-1) ((〖994.8〗^2 〖+1231〗^2-〖753.1〗^2)/(2×994.8×1231))
∠C=37°42'4″

The sine rule formula was used to calculate ∠CBDand∠CDB:
a/(sin A)=b/sinB=c/sinC
BD/(sin C)=CD/(sin∠CBD)
753.1/(sin37°42'4″)=994.8/(sin∠CBD)
∠CBD=sin^(-1) （(994.8×sin⁡〖37°〖42〗^' 4^″ 〗)/753.1）
∠CBD=53°〖52〗^' 57″

BD/(sin C)=CB/(sin∠CDB)
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