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Verify the angles:
∠CDB+∠CBD+∠C=88°〖24〗^' 〖22〗^″+53°〖52〗^' 〖57〗^″+37°〖42〗^' 4^″=179°〖59〗^' 〖23〗^″≈180°
In△CBD，the three angles can be added to 180°. 
Q4. Find the area formed by the lines joining these towns from the above diagram.

The surface areaformula was used:
SA = 1/2 bc×sinA
SA ∆ABD=1/2×AB×AD×sin∠BAD
=1/2×978.3×1379.0×sin31°51'50″
=356090.17〖km〗^2

SA ∆CBD=1/2×CB×CD×sin∠BCD
=1/2×994.8×1231×sin37°42'4″
=374447.03〖km〗^2

The area of triangle ABD is 356090.17km²，the area of triangle CBD is 374447.03Km².

Q5. Verify the bearing using trigonometry and the latitude and longitude values from the diagram below.

∠FAD=∠ADB=43°〖17〗^' 〖44〗^″
∠EAB=∠EAF+∠FAD+∠BAD=90°+43°〖17〗^' 〖44〗^″+31°〖51〗^' 〖50〗^″
=165°9'34″
The initial bearing of Yi Chang to Shang Hai calculated by using latitude and longitude is 162°22′04″, and the trigonometry calculation is 165°9'34″
 
Q6. Write any assumptions, strengths and/or limitations relating to the calculations above.
Assumptions:
 It wasassumed to happenin 2 dimension not 3 dimensions. The earth should be a sphere, but it consider as a flat surface. The distance between the towns and the cyclone is a straight line not curved line.
 It’s was assumed that the centre of the towns, cities and cyclone are joined together.
 The calculations which were provided by the websites were assumed to be accurate.
Limitations:
 The calculations of the questions were not very accurate, they were approximate value.For example, most of the angles were approximate to the neatest seconds.
 The distance between each town and cityfrom the websites may not be accurate.
 

Q7. Find the distance from the weather to each town when the angle to the other town has reduced by a half.
 

In the diagram above, the cyclone was assumed to move directly to the town Yi Chang on AD. And when the cyclone moved to D’, the angle BCD was divided in half by CD’. The cyclone D’ and each town are joined together, A, E, D’, D are on the same line.
According to Q2,
∠A=31°〖51〗^' 〖50〗^″,∠ADB=43°〖17〗^' 〖44〗^″,∠ABD=104°〖50〗^' 26″,
∠C=37°42'4″, ∠CBD=53°〖52〗^' 〖57〗^″, ∠CDB=88°〖24〗^' 22″

∵CD’ divided ∠BDC in half
∴∠BCD^'=∠DCD^'=1/2×∠BDC=1/2×37°〖42〗^' 4^″=18°51'2″
∠CDD^'=∠CDB-∠ADB=88°〖24〗^' 〖22〗^″-43°〖17〗^' 〖44〗^″=45°6'38″
In △CDD’, ∠DCD'=18°〖51〗^' 2^″, ∠CDD'=45°6^' 〖38〗^″, CD=994.8km
∠CD^' D=180°-∠DCD^'-∠CDD^'=180°-18°〖51〗^' 2^″-45°6^' 〖38〗^″=116°2'20″
The sine rule formula was used:
a/(sin A)=b/sinB=c/sinC
CD'/(sin ∠CDD')=CD/(sin∠CD'D)
CD'/(sin45°6^' 〖38〗^″ )=753.1/(sin116°2'20″)
CD'=(753.1×sin45°6^' 〖38〗^″)/(sin116°2'20″)
CD^'=784.41km
DD'/(sin ∠DCD')=CD/(sin∠CD'D)
DD'/(sin18°51'2″)=753.1/(sin116°2'20″)
DD'=(753.1×sin18°51'2″)/(sin116°2'20″)
DD^'=270.81km
〖AD^'=AD-DD〗^'=1379-270.81=1108.19km
In △CBD’, ∠DCD'=18°〖51〗^' 2^″, CD’=784.41Km, CB=1231km
The cosine rule formula was used
a^2=b^2+c^2-2bccosA
〖BD^'〗^2=〖CB〗^2+〖CD^'〗^2-2×CB×CD^'×cos∠BCD'
〖BD^'〗^2=〖1231〗^2+〖784.41〗^2-2×1231×784.41×cos18°〖51〗^' 2^本论文由英语论文网提供整理，提供论文代写，英语论文代写，代写论文，代写英语论文，代写留学生论文，代写英文论文，留学生论文代写相关核心关键词搜索。