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A tower \(B T\) has height \(h\) metres.
From point \(A\), the angle of elevation to the top of the tower is 26° as shown.
Which of the following is the correct expression for the length of \(A B\) ?
For the angle `theta, tan theta = -5/12` and `cos theta = 12/13.`
Which diagram best shows the angle `theta?`
A student is vaping behind a wall in a school assembly room.
A teacher is walking up a corridor as shown in the diagram. The wall at the end is covered by a large mirror.
The line of sight off a mirror follows the law of reflection as follows:
Determine the distance, \(d\) metres, at which the teacher will first be able to see the student vaping. (4 marks)
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\(\text{Label alternate angles at } d_{\text{min}}:\)
\(\tan \theta=\dfrac{x}{2} \ \ \text{(bottom left \(\Delta\))}\)
\(\tan \theta=\dfrac{1-x}{1}\ \ \text{\((\Delta\) with side \(1-x\))}\)
\(\dfrac{x}{2}=1-x\)
\(x=2-2 x\)
\(x=\dfrac{2}{3}\)
\(\text{Consider top left triangle:}\)
| \(\tan \theta\) | \(=\dfrac{\frac{4}{3}}{d}\) |
| \(\dfrac{1}{3}\) | \(=\dfrac{4}{3 d}\) |
| \(d\) | \(=4.0 \ \text{metres}\) |
The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.
The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.
What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place. (4 marks)
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`text{Arc}\ PQ=110/360 xx pi xx 2 xx 2.1=4.03171… \ text{m}`
`text{Consider}\ ΔOPQ:`
| `sin 55^@` | `=x/2.1` | |
| `x` | `=2.1 xx sin 55^@=1.7202…` |
`PQ=2x=3.440\ text{m}`
| `:.\ text{Perimeter}` | `=8+(2xx3.6)+4.031+(8-3.440)` | |
| `=23.79…\ text{m}` | ||
| `=23.8\ text{m (to 1 d.p.)}` |
A right-angled triangle `XYZ` is cut out from a semicircle with centre `O`. The length of the diameter `XZ` is 16 cm and `angle YXZ` = 30°, as shown on the diagram.
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For the angle `theta, sin theta = 7/25` and `cos theta = -24/25.`
Which diagram best shows the angle `theta?`
An advertising logo is formed from two circles, which intersect as shown in the diagram.
The circles intersect at `A` and `B` and have centres at `O` and `C`.
The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.
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i. |
 |
`text(In)\ Delta AOC:`
`AO^2 + AC^2= 1^2 + sqrt 3^2= 4= OC^2`
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`
ii. `sin /_ACO= 1/2\ \ =>\ \ /_ACO= pi/6`
`sin\ /_AOC= sqrt 3/2\ \ =>\ \ /_AOC= pi/3`
iii. `text(Area)\ AOBC`
`= 2 xx text(Area)\ Delta AOC`
`= 2 xx 1/2 xx b xx h`
`= 2 xx 1/2 xx 1 xx sqrt 3`
`= sqrt 3\ \ text(m)^2`
iv. `/_ACB = pi/6 + pi/6 = pi/3`
`/_ACB\ text{(reflex)}= 2 pi-pi/3= (5 pi)/3`
`text(Area of major sector)\ ACB`
`= theta/(2 pi) xx pi r^2`
`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`
`= (5 pi)/6 xx 3`
`= (5 pi)/2\ text(m)^2`
v. `/_AOB = pi/3 + pi/3 = (2 pi)/3`
`/_AOB\ text{(reflex)}= 2 pi-(2 pi)/3= (4 pi)/3`
`text(Area of major sector)\ AOB`
`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`
`= (2 pi)/3\ text(m)^2`
`:.\ text(Total area of the logo)`
`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`
`= (15 pi + 4 pi)/6 + sqrt 3`
`= ((19 pi + 6 sqrt 3)/6)\ text(m)^2`
The triangle `ABC` has a right angle at `B, \ ∠BAC = theta` and `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.
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| a. |  |
`text(Show)\ EF = 6 sin^3 theta`
`text(In)\ ΔADB:`
| `sin theta` | `= (DB)/6` |
| `DB` | `= 6 sin theta` |
| `∠ABD` | `= 90-theta\ \ \ text{(angle sum of}\ ΔADB)` |
| `:.∠DBE` | `= theta\ \ \ (∠ABE\ text{is a right angle)}` |
`text(In)\ ΔBDE:`
`sin theta= (DE)/(DB)= (DE)/(6 sin theta)`
| `DE` | `= 6 sin^2 theta` |
| `∠BDE` | `= 90-theta\ \ \ text{(angle sum of}\ ΔDBE)` |
| `∠EDF` | `= theta\ \ \ (∠FDB\ text{is a right angle)}` |
`text(In)\ ΔDEF:`
| `sin theta` | `= (EF)/(DE)= (EF)/(6 sin^2 theta)` |
| `:.EF` | `= 6 sin^3 theta\ \ …text(as required)` |
b. `text(Show)\ \ BD + EF + GH\ …`
`text(has limiting sum)\ =6 sec theta tan theta`
`underbrace{6 sin theta + 6 sin^3 theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
`text(S)text(ince)\ \ 0 < theta < 90^@`
| `-1` | `< sin\ theta` | `< 1` |
| `0` | `< sin^2\ theta` | `< 1` |
`:. |\ r\ | < 1`
| `:.S_∞` | `= a/(1-r)` |
| `= (6 sin theta)/(1-sin^2 theta)` | |
| `= (6 sin theta)/(cos^2 theta)` | |
| `= 6 xx 1/(cos theta) xx (sin theta)/(cos theta)` | |
| `= 6 sec theta tan theta\ \ …text( as required.)` |
The diagram shows the line `l`.
What is the slope of the line `l`?