The diagram shows triangle `ABC`.
Calculate the area of the triangle, to the nearest square metre. (3 marks)
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The diagram shows triangle `ABC`.
Calculate the area of the triangle, to the nearest square metre. (3 marks)
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`147\ text{m}^2`
`text{Using the sine rule:}`
`(CB)/sin60^@` | `=12/sin25^@` | |
`CB` | `=sin60^@ xx 12/sin25^@` | |
`=24.590…` |
`angleACB=180-(60+25)=95^@\ \ text{(180° in Δ)}`
`text{Using the sine rule (Area):}`
`A` | `=1/2 xx AC xx CB xx sin angleACB` | |
`=1/2 xx 12 xx 24.59 xx sin95^@` | ||
`=146.98…` | ||
`=147\ text{m}^2` |
The diagram shows two right-angled triangles, `ABC` and `ABD`,
where `AC=35 \ text{cm},BD=93 \ text{cm}, /_ACB=41^(@)` and `/_ADB=theta`.
Calculate the size of angle `theta`, to the nearest minute. (4 marks)
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`19^@6^{′}`
`text{In}\ Delta ABC:`
`cos 41^@` | `=35/(BC)` | |
`BC` | `=35/(cos 41^@)` | |
`=46.375…` |
`angle BCD = 180-41=139^@`
`text{Using sine rule in}\ Delta BCD:`
`sin theta/(46.375)` | `=sin139^@/93` | |
`sin theta` | `=(sin 139^@ xx 46.375)/93` | |
`:.theta` | `=sin^(-1)((sin 139^@ xx 46.375)/93)` | |
`=19.09…` | ||
`=19^@6^{′}\ \ text{(nearest minute)}` |
The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.
Find the size of the obtuse angle `ABC` correct to the nearest degree. (3 marks)
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`133°`
`text(Using the sine rule:)`
`sin theta/25` | `= (sin 28°)/16` |
`sin theta` | `= (25 xx sin 28°)/16` |
`sin theta` | `= 0.73355` |
`theta` | `= 47°` |
`:. angleABC` | `= 180-47` |
`= 133°` |
The diagram shows a triangle with side lengths 25 cm and 47 cm and angle 30° and `theta`.
Find `theta` given it is an obtuse angle. Give your answer to the nearest minute. (3 marks)
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`109°57′`
`text(Using sine rule:)`
`(sintheta)/47` | `= (sin30°)/25` |
`sintheta` | `=47 xx (1/2)/25` |
`sintheta` | `= 47/50` |
`theta` | `= sin^(−1)\ 47/50` |
`= 70.05\ text(or)\ 109.94` | |
`= 109.948…\ \ (theta\ text(is obtuse))` | |
`= 109°57′` |
Which formula should be used to calculate the distance between Toby and Frankie?
`A`
`text(The triangle is not a right-angled triangle,)`
`:.\ text(Not)\ B`
`text(Given the information on the diagram provides)`
`text(2 angles and 1 side, the sine rule will work best.)`
`a/sinA = b/sinB`
`=> A`
The angle of depression from `J` to `M` is 75°. The length of `JK` is 20 m and the length of `MK` is 18 m.
Copy or trace this diagram into your writing booklet and calculate the angle of elevation from `M` to `K`. Give your answer to the nearest degree. (3 marks)
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`58^@`
`/_AJL = 90^@`
`/_MJL = 90 – 75 = 15^@`
`text(Using sine rule in)\ Delta MJK`
`text(Let)\ /_JMK = x^@`
`20/sin x` | `= 18/sin15^@` |
`18 sin x` | `= 20 xx sin 15^@` |
`sin x` | `= (20 xx sin 15^@)/18 = 0.2875…` |
`x` | `= 16.71…^@` |
`/_JML = 75^@\ text{(} text(alternate angles,)\ ML \ text(||) \ AJ text{)}` |
`:.\ /_KML` | `= 75^@ – 16.71…` |
`= 58.287…^@` | |
`= 58^@\ \ \ text{(nearest degree)}` |
`:.\ text(Angle of Elevation from)\ M\ text(to)\ K\ text(is)\ 58^@.`
Find the area of triangle `ABC`, correct to the nearest square metre. (3 marks)
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`717\ text(m²)` `text{(nearest m²)}`
`cos/_C` | `=(AC^2 + CB^2-AB^2)/(2 xx AC xx CB)` |
`=(50^2 + 40^2-83^2)/(2 xx 50 xx 40)` | |
`= -0.69725…` | |
`/_C` | `=134.2067…^@` |
`text(Using Area) = 1/2 ab\ sinC :` |
`text(Area)\ Delta ABC` | `=1/2 xx 50 xx 40 xx sin134.2067…^@` |
`=716.828…` | |
`=717\ text(m²)\ \ \ \ text{(nearest m²)}` |
In the diagram, `AD` and `DC` are equal to 30 cm.
What is the length of `AB` to the nearest centimetre?
(A) `28\ text(cm)`
(B) `31\ text(cm)`
(C) `34\ text(cm)`
(D) `39\ text(cm)`
`A`
`Delta ADC\ text(is isosceles)`
`/_DAB = /_DCA` | `= x^@` |
`2x + 80^@` | `= 180^@\ \ \ (text{Angle sum of}\ DeltaADC)` |
`2x` | `= 100^@` |
`x` | `= 50^@` |
`/_ DBA` | `= 180\-(50 + 60)\ \ \ (text{Angle sum of}\ Delta ADB)` |
`= 70^@` |
`text(Using sine rule:)`
`(AB)/sin60` | `= 30/sin70` |
`AB` | `= (30 xx sin60)/sin70` |
`= 27.648…\ text(cm)` |
`=> A`
Three towns `P`, `Q` and `R` are marked on the diagram.
The distance from `R` to `P` is 76 km. `angle RQP=26^circ` and `angle RPQ=46^@`
What is the distance from `P` to `Q` to the nearest kilometre?
`C`
`angle QRP` | `=180-(26+46) (180^circ\ text(in) \ Delta)` |
`=108^circ` |
`text{Using sine rule}`
`(PQ)/sin108^circ` | `=76/sin26^circ` |
`PQ` | `=(76xxsin108^circ)/sin26^circ` |
`=164.88\ text(km)` |
`=> C`
Raj cycles around a course. The course starts at `E`, passes through `F`, `G` and `H` and finishes at `E`. The distances `EH` and `GH` are equal.
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i. `text(Find)\ EF:`
`/_ FGE` | `= 180\-(139 + 31) \ \ text{(angle sum of}\ Delta EFGtext{)}` |
`= 180-170` | |
`= 10^@` |
`text(Using Sine rule:)`
`(EF)/sin10^@` | `= 82/sin139^@` |
`EF` | `= (82 xx sin10^@)/sin139^@` |
`= 21.70406…` | |
`= 22\ text{km (nearest km)}` |
ii. `text(Let)\ \ d = text(total distance cycled)`
`text(Find)\ EH`
`text(S)text(ince)\ Delta EGH\ text(is isosceles, and)\ /_EHG = 90^@`
`/_GEH = /_HGE = 45^@`
`text{(angles opposite equal sides in}\ Delta EGHtext{)}`
`sin45^@` | `= (GH)/82` |
`GH` | `= 82 xx sin45^@` |
`= 57.983…` |
`:. d` | `= EF + FG + GH + EH` |
`= 21.704… + 64 + 57.983 + 57.983` | |
`= 201.66…` | |
`= 202 \ text{km (nearest km)}` |
`C`
`a/sinA` | `=b/sinB` |
`82/sinA` | `=100/sin26` |
`sin A` | `=(82 xx sin26)/100` |
`=0.35946…` | |
`/_A` | `=21^@\ \ \ \ text{(nearest degree)}` |
`text(S)text(ince)\ 180^@\ text(in)\ Delta:`
`90+26+(theta+21)` | `=180` |
`theta` | `=43^@` |
`=> C`