Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up

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Presentation transcript:

Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C 27,800 30,500 33,200 36,000 B Find the area of Quadrilateral A B C D, in feet squared. [pause] In this problem, --- N A D

Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C 27,800 30,500 33,200 36,000 B survey data is provided, where a total station was set up at Point A, --- N A D

Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C 27,800 30,500 33,200 36,000 B And the bearings and lengths were measured, when sighting Points B, ---- N A D

Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C 27,800 30,500 33,200 36,000 B C and D. [pause] To find the area inside Quadrilateral A B C D, we’ll divide the area into ---- N A D

Find: AreaABCD [ft2] C 27,800 30,500 B 33,200 36,000 N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C 27,800 30,500 33,200 36,000 B 2 triangles. So we’ll have an area for Triangle A B C, and an area for --- N A D

Find: AreaABCD [ft2] C B AreaABC AreaACD N A D set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C B Triangle A C D. Where the sum of the areas from these triangles, equals, --- AreaABC AreaACD N A D

+ Find: AreaABCD [ft2] C AreaABC AreaACD B AreaABC AreaABCD AreaACD N set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC + AreaACD B the area of the quadrilateral. If we focus in on just Triangle A B C, --- AreaABC AreaABCD AreaACD N A D

+ Find: AreaABCD [ft2] C AreaABC AreaACD B AreaABC AreaABCD AreaACD N set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC + AreaACD B we’ll solve for this area using the equation, area equals ---- AreaABC AreaABCD AreaACD N A D

+ Find: AreaABCD [ft2] C AreaABC AreaACD B AreaABC AreaABCD set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC + AreaACD B one half times side length A B, times side length A C, time the sine of angle B A C. The survey data provides --- AreaABC AreaABCD AreaABC AABC = 0.5 * LAB * LAC * sin(ABAC) N A

+ Find: AreaABCD [ft2] C AreaABC AreaACD B AreaABC AreaABCD set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC + AreaACD B Length A B and Length A C, as 152 feet and 237 feet, respectively. Angle B A C --- AreaABC AreaABCD AreaABC AABC = 0.5 * LAB * LAC * sin(ABAC) N A

+ Find: AreaABCD [ft2] C AreaABC AreaABC AreaACD B ABAC AreaABCD set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC AreaABC + AreaACD B can be determined since we know the bearing from Point A to Point B and --- ABAC AreaABCD AABC = 0.5 * LAB * LAC * sin(ABAC) N AreaABC A

+ Find: AreaABCD [ft2] C AreaABC AreaABC AreaACD B ABAC AreaABCD set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC AreaABC + AreaACD B the bearing from Point A to Point C. From these bearing measurements, --- ABAC AreaABCD AABC = 0.5 * LAB * LAC * sin(ABAC) N AreaABC A

Find: AreaABCD [ft2] C AreaABC B ABAC course bearing azimuth o AB N40 30’ 00”W AzAB AC N52 15’ 00”E o AzAC o AD S74 45’ 00”E AzAD C AreaABC B we can determine the azimuth values … and then equate angle B A C to the the Azimuth of --- 152 [ft] ABAC 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) N AreaABC A

Find: AreaABCD [ft2] C AreaABC B ABAC course bearing azimuth o AB N40 30’ 00”W AzAB AC N52 15’ 00”E o AzAC o AD S74 45’ 00”E AzAD C AreaABC B course A C, minus, the azimuth of course A B, plus 360 degrees. The azimuth of a certain direction is the angle ---- 152 [ft] ABAC 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) N AreaABC o A AzAC-AzAB+360

Find: AreaABCD [ft2] C N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC course bearing azimuth o AB N40 30’ 00”W AzAB o AC N52 15’ 00”E AzAC o AD S74 45’ 00”E AzAD C N AzAC B an observer must turn from due north, in a clockwise direction, to be aligned, with that direction. Therefore, Azimuth A C is simply, --- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] C N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC course bearing azimuth o AB N40 30’ 00”W AzAB o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C N AzAC B the same measurement as the bearing, 52.25 degrees, since course A C runs in the northeast direction, --- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] C N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC course bearing azimuth o AB N40 30’ 00”W AzAB o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C N AzAC B which is in quadrant 1. The azimuth of course A B is ---- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC A course bearing azimuth o AB N40 30’ 00”W AzAB o AC N52 15’ 00”E o 52.25 o AD S74 45’ 00”E AzAD o o AzAB=360 -40 30’ 00” N AzAC B 360 degrees minus, 40 degrees 30 minutes, which equals --- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC A course bearing azimuth o o AB N40 30’ 00”W 319.50 o AC N52 15’ 00”E o 52.25 o AD S74 45’ 00”E AzAD o o AzAB=360 -40 30’ 00” N o AzAB=319.50 AzAC B 319.50 degrees. [pause] After plugging in---- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC A course bearing azimuth o o AB N40 30’ 00”W 319.50 o AC N52 15’ 00”E o 52.25 o AD S74 45’ 00”E AzAD o o AzAB=360 -40 30’ 00” N o AzAB=319.50 AzAC B these two azimuth values, Angle B A C equals ---- 152 [ft] 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360

Find: AreaABCD [ft2] C N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC course bearing azimuth o o AB N40 30’ 00”W 319.50 N52 15’ 00”E o o AC 52.25 o AD S74 45’ 00”E AzAD C N AzAC B 92.75 degrees. [pause] And after plugging in the given lengths --- 152 [ft] 237 [ft] ABAC AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360 o 92.75

Find: AreaABCD [ft2] C N B AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC course bearing azimuth o o AB N40 30’ 00”W 319.50 N52 15’ 00”E o o AC 52.25 o AD S74 45’ 00”E AzAD C N AzAC B L A B and L A C, Area A B C equals, --- 152 [ft] 237 [ft] ABAC AABC = 0.5 * LAB * LAC * sin(ABAC) AreaABC o AzAB A AzAC-AzAB+360 o 92.75

Find: AreaABCD [ft2] C AreaABC AreaABC=17,991[ft2] B ABAC course bearing azimuth o o AB N40 30’ 00”W 319.50 N52 15’ 00”E o o AC 52.25 o AD S74 45’ 00”E AzAD C AreaABC AreaABC=17,991[ft2] B 17,991 feet squared. [pause] Knowing the area of Triangle A B C, --- 152 [ft] ABAC 237 [ft] AABC = 0.5 * LAB * LAC * sin(ABAC) N AreaABC o A AzAC-AzAB+360 o 92.75

+ Find: AreaABCD [ft2] C AreaABC=17,991[ft2] AreaABC B AreaABC ABAC course bearing azimuth o o AB N40 30’ 00”W 319.50 N52 15’ 00”E o o AC 52.25 o AD S74 45’ 00”E AzAD C AreaABC AreaABC=17,991[ft2] B AreaABC We still need to determine the area of Triangle, --- ABAC + AreaACD AreaABCD N A

+ Find: AreaABCD [ft2] C AreaABC=17,991[ft2] AreaABC B AreaABC AreaACD course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C AreaABC AreaABC=17,991[ft2] B AreaABC A C D. [pause] We’ll solve for the area of Triangle A C D, --- + AreaACD AreaABCD N A D

+ Find: AreaABCD [ft2] C AreaABC=17,991[ft2] AreaABC AreaACD AreaABCD course bearing azimuth o o AB N40 30’ 00”W 319.50 o AC N52 15’ 00”E o 52.25 o AD S74 45’ 00”E AzAD C AreaABC=17,991[ft2] AreaABC using the same equation as before, where the area of Triangle A C D equals, --- + AreaACD AreaABCD N A D

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N AreaABC set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC=17,991[ft2] N AreaABC one half times the length of side A C, times the length of side A D, times the sine of Angle C A D. AreaACD=0.5*LAC*LAD*sin(ACAD) A D

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N set up sighting bearing length [ft] o A B N40 30’ 00”W 152 o A C N52 15’ 00”E 237 o A D S74 45’ 00”E 161 C AreaABC=17,991[ft2] N Lengths A C and A D are provided in the survey data … and we’ll calculate Angle C A D, --- AreaACD=0.5*LAC*LAD*sin(ACAD) A D

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] by subtracting the Azimuth of course A C from the azimuth of course A D. From earlier, --- AreaACD=0.5*LAC*LAD*sin(ACAD) ACAD ACAD=AzAD-AzAC A D

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] we already calculated the azimuth of course A C, as, 52.25 degrees. The azimuth of course A D equals, --- AreaACD=0.5*LAC*LAD*sin(ACAD) ACAD ACAD=AzAD-AzAC A D

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o AD S74 45’ 00”E AzAD C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] 180 degrees minus, 74 degrees 45 minutes, which equates to, --- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD ACAD ACAD=AzAD-AzAC A o o D 180- 74 45’ 00”

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] 105.25 degrees. This makes Angle C A D equal to --- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD ACAD ACAD=AzAD-AzAC A o o o D 105.25 = 180- 74 45’ 00”

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] 105.25 degrees minus 52.25 degrees, which equals --- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD ACAD ACAD=AzAD-AzAC A o o o D 105.25 = 180- 74 45’ 00”

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] 53.00 degrees. Plugging 53 degrees in for Angle C A D, as well as --- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD o 53.00 ACAD ACAD=AzAD-AzAC A o o o D 105.25 = 180- 74 45’ 00”

Find: AreaABCD [ft2] C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 C AreaABC=17,991[ft2] N 237 [ft] 161 [ft] the values for Length A C and A D, Area A C D equals, --- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD o 53.00 o ACAD 52.25 ACAD=AzAD-AzAC A o o o D 105.25 = 180- 74 45’ 00”

Find: AreaABCD [ft2] AreaACD = 15,237 [ft2] AreaABC=17,991[ft2] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 AreaACD = 15,237 [ft2] AreaABC=17,991[ft2] 237 [ft] 161 [ft] 15,237 feet squared. [pause] The area of quadrilateral A B C D, equals, ---- AreaACD=0.5*LAC*LAD*sin(ACAD) AzAD o 53.00 o ACAD 52.25 ACAD=AzAD-AzAC A o o o D 105.25 = 180- 74 45’ 00”

+ Find: AreaABCD [ft2] AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] AreaABC 17,991 feet squared, plus 15,237, feet squared, which equals, --- AzAD + AreaACD ACAD AreaABCD A D

+ Find: AreaABCD [ft2] AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] AreaABC 33,228 feet squared. [pause] AzAD + AreaACD ACAD AreaABCD = 33,228 [ft2] A D

+ Find: AreaABCD [ft2] AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] 27,800 30,500 33,200 36,000 AreaABC After looking over the possible solutions, --- + AreaACD AreaABCD = 33,228 [ft2]

+ Find: AreaABCD [ft2] AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] course bearing azimuth o o AB N40 30’ 00”W 319.50 o o AC N52 15’ 00”E 52.25 o o AD S74 45’ 00”E 105.25 AreaABC=17,991[ft2] AreaACD = 15,237 [ft2] 27,800 30,500 33,200 36,000 AreaABC the answer is C. + AreaACD AreaABCD = 33,228 [ft2] AnswerC

Find the area of Triangle A B C, in square feet Find the area of Triangle A B C, in square feet. [pause] In this problem, ----