+26 
The diagram above shows the junction of two rivers. Some scouts are asked to estimate the distance between the point C and the tree, T, on the opposite bank. The scouts are at the point B and cannot cross either river. They mark the points A, B and D and make the measurements shown in the diagram. ABC and DBT are straight lines.
AB = 20 m, BD = 15 m, AD = 17 m, angle TAB = 41° and angle BDC = 36°.
Using a scale of 1 centimetre to represent 5 metres, draw a scale diagram of the situation and find the distance CT, in metres.
I am actually struggling with this one....>.<
+128475 I can see why, Panda....it is a little sticky!!
First, let's use the Law of Cosines to find angle ABD...so we have
17^2 = 15^2 + 20^2 - 2(15)(20)cosABD
289 = 225 + 400 - 600cosABD
cosABD = [289-225-400] / (-600)
cos-1 (336 / 600) = about 55.9 degrees
And because they are vertical angles, this is the same measure as TBC
And DBC is supplemental = 124.1 degrees
And BCD = (180 - 124.1 - 36) = 19.9 degrees
And we can find BC using the Law of Sines
BC/sin36 = 15/sinBCD
BC/sin36 = 15/sin19.9
BC = 15sin36/sin19.9 = about 25.9
And, because the are vertical angles, ABT = DBC = 124.1 degrees
And ATB = (180 - 41 - 124.1) = 14.9 degrees
And we can find BT using the Law of Sines
BT/sin41 = 20/sinATB
BT/sin41 =20/sin14.9
BT = 20sin41/sin14.9 = about 51
And TBC = ABD = 55.9
And finally, using the Law of Cosines one more time, we can find CT
CT^2 = BT^2 + BC^2 - 2(BT)(BC)cosABD
CT^2 = 51^2 + 25.9^2 - 2(51)(25.9)cos(55.9)
CT = √[51^2 + 25.9^2 - 2(51)(25.9)cos(55.9)]= about 42.32 m
Whew!!!...that one was kinda' tough !!!
+128475 I can see why, Panda....it is a little sticky!!
First, let's use the Law of Cosines to find angle ABD...so we have
17^2 = 15^2 + 20^2 - 2(15)(20)cosABD
289 = 225 + 400 - 600cosABD
cosABD = [289-225-400] / (-600)
cos-1 (336 / 600) = about 55.9 degrees
And because they are vertical angles, this is the same measure as TBC
And DBC is supplemental = 124.1 degrees
And BCD = (180 - 124.1 - 36) = 19.9 degrees
And we can find BC using the Law of Sines
BC/sin36 = 15/sinBCD
BC/sin36 = 15/sin19.9
BC = 15sin36/sin19.9 = about 25.9
And, because the are vertical angles, ABT = DBC = 124.1 degrees
And ATB = (180 - 41 - 124.1) = 14.9 degrees
And we can find BT using the Law of Sines
BT/sin41 = 20/sinATB
BT/sin41 =20/sin14.9
BT = 20sin41/sin14.9 = about 51
And TBC = ABD = 55.9
And finally, using the Law of Cosines one more time, we can find CT
CT^2 = BT^2 + BC^2 - 2(BT)(BC)cosABD
CT^2 = 51^2 + 25.9^2 - 2(51)(25.9)cos(55.9)
CT = √[51^2 + 25.9^2 - 2(51)(25.9)cos(55.9)]= about 42.32 m
Whew!!!...that one was kinda' tough !!!
