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# Please someone help me. Thanks.

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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....>.<

Feb 4, 2015

#1
+94619
+15

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 !!!

Feb 4, 2015

#1
+94619
+15

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 !!!

CPhill Feb 4, 2015
#2
+1000
0

i was so stumped! XD dont blame me im only in seventh grade!

Feb 4, 2015
#3
+26
0

Thank you CPhill! It is still a bit confusing, but thanks for the help, it helped me. c:

Feb 4, 2015
#4
+95369
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Very impressive Chris :)

Feb 4, 2015
#5
+94619
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Thanks, Mel.....!!!

Feb 4, 2015