Consider the diagram below.
What is the true bearing of `A` from `B`?
- `025^@`
- `065^@`
- `115^@`
- `295^@`
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Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.
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a. | `angle APB` | `= 100 – 35` |
`= 65^@` |
b. `text(Using cosine rule:)`
`AB^2` | `= AP^2 + PB^2 – 2 xx AP xx PB cos 65^@` |
`= 49 + 81 – 2 xx 7 xx 9 cos 65^@` | |
`= 76.750…` | |
`:.AB` | `= 8.760…` |
`= 8.76\ text{km (to 2 d.p.)}` |
c.
`anglePAC = 35^@\ (text(alternate))`
`text(Using cosine rule, find)\ anglePAB:`
`cos anglePAB` | `= (7^2 + 8.76 – 9^2)/(2 xx 7 xx 8.76)` | |
`= 0.3647…` | ||
`:. angle PAB` | `= 68.61…^@` | |
`= 69^@\ \ (text(nearest degree))` |
`:. text(Bearing of)\ B\ text(from)\ A\ (theta)`
`= 180 – (69 – 35)`
`= 146^@`
A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`. The
size of angle `ABC` is 114°.
Copy the diagram into your writing booklet and show all the information on it.
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i. |
`text(Let point)\ D\ text(be due North of point)\ B`
`/_ABD` | `=180-121\ text{(cointerior with}\ \ /_A text{)}` |
`=59^@` | |
`/_DBC` | `=114-59` |
`=55^@` |
`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`
ii. `text(Using cosine rule:)`
`AC^2` | `=AB^2+BC^2-2xxABxxBCxxcos/_ABC` |
`=6^2+9^2-2xx6xx9xxcos114^@` | |
`=160.9275…` | |
`:.AC` | `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}` |
`=13\ text(km)\ text{(nearest km)}` |
iii. `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`
`cos/_ACB` | `=(AC^2+BC^2-AB^2)/(2xxACxxBC)` |
`=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)` | |
`=0.9018…` | |
`/_ACB` | `=25.6^@\ text{(to 1 d.p.)}` |
`text(From diagram,)`
`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`
`:.\ text(Bearing of)\ A\ text(from)\ C`
`=180+55+25.6` | |
`=260.6` | |
`=261^@\ text{(nearest degree)}` |
The diagram shows the location of three schools. School `A` is 5 km due north of school `B`, school `C` is 13 km from school `B` and `angleABC` is 135°.
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(i) `text(Using cosine rule:)`
`AC^2` | `= AB^2 + BC^2 – 2 xx AB xx BC xx cos135^@` |
`= 5^2 + 13^2 – 2 xx 5 xx 13 xx cos135^@` | |
`= 285.923…` |
`:. AC` | `= 16.909…` |
`= 17\ text{km (nearest km)}` |
(ii) |
`text(Using sine rule, find)\ angleBAC:`
`(sin angleBAC)/13` | `= (sin 135^@)/17` |
`sin angleBAC` | `= (13 xx sin 135^@)/17` |
`= 0.5407…` | |
`angleBAC` | `= 32.7^@` |
`:. text(Bearing of)\ C\ text(from)\ A`
`= 180 + 32.7`
`= 212.7^@`
`= 213^@`
The bearing of `C` from `A` is 250° and the distance of `C` from `A` is 36 km.
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i. `text(There is 360)^@\ text(about point)\ A`
`:.theta + 250^@` | `= 360^@` |
`theta` | `= 110^@` |
ii. |
`a^2` | `= b^2 + c^2 − 2ab\ cos\ A` |
`CB^2` | `= 36^2 + 15^2 − 2 xx 36 xx 15 xx cos\ 110^@` |
`= 1296 + 225 −(text(−369.38…))` | |
`= 1890.38…` | |
`:.CB` | `= 43.47…` |
`= 43\ text{km (nearest km)}` |
The diagram shows information about the locations of towns `A`, `B` and `Q`.
Calculate her walking speed correct to the nearest km/h. (1 mark)
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Find the distance from Town `A` to Town `B`. Give your answer to the nearest km. (2 marks)
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i. `text(2 hrs 48 mins) = 168\ text(mins)`
`text(Speed)\ text{(} A\ text(to)\ Q text{)}` | `= 15/168` |
`= 0.0892…\ text(km/min)` |
`text(Speed)\ text{(in km/hr)}` | `= 0.0892… xx 60` |
`= 5.357…\ text(km/hr)` | |
`= 5\ text(km/hr)\ text{(nearest km/hr)}` |
ii. |
`text(Using cosine rule)`
`AB^2` | `= 15^2 + 10^2 – 2 xx 15 xx 10 xx cos 87^@` |
`= 309.299…` | |
`AB` | `= 17.586…` |
`= 18\ text(km)\ text{(nearest km)}` |
`:.\ text(The distance from Town)\ A\ text(to Town)\ B\ text(is 18 km.)`
iii. |
`/_CAQ` | `= 31^@\ \ \ text{(} text(straight angle at)\ A text{)}` |
`/_AQD` | `= 31^@\ \ \ text{(} text(alternate angle)\ AC\ text(||)\ DQ text{)}` |
`/_DQB` | `= 87 – 31 = 56^@` |
`/_QBE` | `= 56^@\ \ \ text{(} text(alternate angle)\ DQ \ text(||)\ BE text{)}` |
`:.\ text(Bearing of)\ Q\ text(from)\ B`
`= 180 + 56`
`= 236^@`
A yacht race follows the triangular course shown in the diagram. The course from `P` to `Q` is 1.8 km on a true bearing of 058°.
At `Q` the course changes direction. The course from `Q` to `R` is 2.7 km and `/_PQR = 74^@`.
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What is the area of this ‘no-go’ zone? (1 mark)
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i. |
`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`
`text(Bearing of)\ R\ text(from)\ Q`
`= 180^@ + 58^@ + 74^@` |
`= 312^@` |
(ii) `text(Using Cosine rule:)`
`RP^2` | `=RQ^2` + `PQ^2` `- 2` `xx RQ` `xx PQ` `xx cos` `/_RQP` |
`= 2.7^2` + `1.8^2` `- 2` `xx 2.7` `xx 1.8` `xx cos74^@` | |
`=7.29 + 3.24\ – 2.679…` | |
`=7.851…` |
`:.RP` | `= sqrt(7.851…)` |
`=2.8019…` | |
`~~ 2.8\ text(km) (text(1 d.p.) )` |
(iii) `text(Using)\ A = 1/2 ab sinC`
`A` | `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@` |
`= 2.3358…` | |
`= 2.3\ text(km²)` |
`:.\ text(No-go zone is 2.3 km²)`
The diagram shows towns `A`, `B` and `C`. Town `B` is 40 km due north of town `A`. The distance from `B` to `C` is 18 km and the bearing of `C` from `A` is 025°. It is known that `∠BCA` is obtuse.
What is the bearing of `C` from `B`?
(A) `070°`
(B) `095°`
(C) `110°`
(D) `135°`
`=> D`
`text(Using the sine rule,)`
`(sin∠BCA)/40` | `= (sin25^@)/18` |
`sin angle BCA` | `= (40 xx sin25^@)/18` |
`= 0.939…` | |
`angle BCA` | `= 180 – 69.9quad(angleBCA > 90^@)` |
`= 110.1°` |
`:. text(Bearing of)\ C\ text(from)\ B`
`= 110.1 + 25qquad(text(external angle of triangle))`
`= 135.1`
`=> D`
The following information is given about the locations of three towns `X`, `Y` and `Z`:
• `X` is due east of `Z`
• `X` is on a bearing of `145^@` from `Y`
• `Y` is on a bearing of `060^@` from `Z`.
Which diagram best represents this information?
`C`
`text(S)text(ince)\ X\ text(is due east of)\ Z`
`=> text(Cannot be)\ B\ text(or)\ D`
`text(The diagram shows we can find)`
`/_ZYX = 60 + 35^@ = 95^@`
`text(Using alternate angles)\ (60^@)\ text(and)`
`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`
`=> C`
The diagram shows the position of `Q`, `R` and `T` relative to `P`.
In the diagram,
`Q` is south-west of `P`
`R` is north-west of `P`
`/_QPT` is 165°
What is the bearing of `T` from `P`?
(A) `060^@`
(B) `075^@`
(C) `105^@`
(D) `120^@`
`A`
Town `B` is 80 km due north of Town `A` and 59 km from Town `C`.
Town `A` is 31 km from Town `C`.
What is the bearing of Town `C` from Town `B`?
(A) `019^@`
(B) `122^@`
(C) `161^@`
(D) `341^@`
`C`
`text(Using the cosine rule:)`
`cos\ /_B` | `= (a^2 + c^2 -b^2)/(2ac)` |
`= (59^2 + 80^2 -31^2)/(2 xx 59 xx 80)` | |
`= 0.9449…` | |
`/_B` | `= 19^@\ text((nearest degree))` |
`:.\ text(Bearing of Town C from B) = 180-19= 161^@`
`=> C`
A ship travels from Port A on a bearing of 050° for 320 km to Port B. It then travels on a bearing of 120° for 190 km to Port C.
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i. |
`text(Let)\ \ D\ \ text(be south of)\ \ B`
`/_ ABD = 50^@ qquad text{(alternate angles)}`
`/_DBC = 60^@ qquad text{(180° in straight line)}`
`:. /_ ABC` | `= 50 + 60` |
`= 110^@` |
ii. `text(Using the cosine rule:)`
`AC^2` | `= AB^2 + BC^2 – 2 *AB*BC* cos /_ ABC` |
`= 320^2 + 190^2 – 2 xx 320 xx 190 xx cos 110^@` | |
`= 180\ 089.64…` | |
`:. AC` | `= 424.36…` |
`= 420\ text{km (nearest 10 km)}` |
The diagram shows a point `P` which is 30 km due west of the point `Q`.
The point `R` is 12 km from `P` and has a bearing from `P` of 070°.
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i. `text(Join)\ \ RQ\ \ text(to form)\ \ Delta RPQ`
`/_RPQ = 90 – 70 = 20^@`
`text(Using the cosine rule:)`
`RQ^2` | `= PR^2 + PQ^2 – 2 xx PR xx PQ xx cos /_RPQ` |
`= 12^2 + 30^2 – 2 xx 12 xx 30 xx cos 20^@` | |
`= 367.421…` |
`:.\ RQ` | `= 19.168…` |
`= 19.2\ text(km)\ \ text{(1 d.p.)}` |
ii. `text(Using sine rule:)`
`(sin /_RQP)/12` | `= (sin 20^@)/(19.168…)` |
`sin/_RQP` | `= (12 xx sin 20^@)/(19.168…)` |
`= 0.214…` | |
`/_RQP` | `= 12.36…^@` |
`= 12^@\ \ \ text{(nearest degree)}` |
`:.\ text(Bearing of)\ R\ text(from)\ Q`
`=270+12`
`=282^@`
Chris leaves island `A` in a boat and sails 142 km on a bearing of 078° to island `B`. Chris then sails on a bearing of 191° for 220 km to island `C`, as shown in the diagram.
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i. |
`text(Find)\ \ /_ABC`
`text(Let)\ D\ text(be south of)\ B`
`:.\ /_CBD = 191\ – 180 = 11°`
`/_ DBA` | `= 78°\ text{(alternate)}` |
`/_ ABC` | `= 78\ – 11` |
`= 67°` |
`text(Using Cosine rule:)`
`AC^2` | `= AB^2 + BC^2\ – 2 * AB * BC * cos /_ABC` |
`= 142^2 + 220^2\ – 2 xx 142 xx 220 xx cos 67°` | |
`= 44\ 151.119…` |
`:.\ AC` | `= 210.121…` |
`~~ 210\ text(km)\ \ \ text(… as required)` |
ii. `text(Find)\ \ /_ ACB`
`text(Using Sine rule:)`
`(sin /_ ACB)/142` | `= (sin /_ABC)/210` |
`sin /_ ACB` | `= (142 xx sin 67°)/210` |
`= 0.6224…` | |
`/_ ACB` | `= 38.494…` |
`= 38°\ text{(nearest degree)}` |
`text(Let)\ E\ text(be due North of)\ C`
`/_ECB = 11°\ text{(} text(alternate to)\ /_CBD text{)}`
`:.\ /_ECA` | `= 38\ – 11` |
`= 27°` |
`:.\ text(Bearing of)\ A\ text(from)\ C`
`= 360\ – 27`
`= 333°`