Detailed Answer

a) \(a = 11.7 * cos(59°)\)

b) \(a = tan^{-1}(\frac{6.5}{10})\)

c) \(b = \frac{7}{sin(35.5°)}\)

d) \(a = cos^{-1}(\frac{3.6}{10.5})\)

e) \(𝛽 = sin^{-1}(\frac{2.5}{9.6})\)

f) \(c = \frac{9.2}{tan(49°)}\)

Detailed Answer

a) With the sine rule: \(\frac{a}{\sin(𝛼)} = \frac{21.4}{\sin(𝛾)}\), so \(a = \sin(43°) \times \frac{21.4}{\sin(104°)} = 15\)

b) We can read the values on the graph or find it in the GDC.

\[ x = -19, \ x=12 \]

c) The law of cosines gives us: \(a^2 + b^2 - 2ab\cos(𝛾) = c^2\). Rearranging the equation gives us \(a^2 + b^2 - 2ab\cos(𝛾) - c^2 = 0\), which we notice is the function we have graphed before. Since \(b\) cannot be negative, we have \(b = 12\).

Detailed Answer

a) We use the formula for the area of a triangle: \(\frac{1}{2} \times 2 \times 2 \sin(60°) = 1.73\,m^2\)

b) We notice that ABC is an isosceles triangle, so angles A and B are equal and equal to: \(𝛼 = 𝛽 = \frac{180° - 60°}{2} = 60°\). This can already tell us that ABC is equilateral, but we can also check this with the sine rule. So, in the end, \(AB = 2\) meters.

c) The semi-circles have a total area of \(\pi \times 1^2 = \pi\). So, the total circle area is \(4.87\,m^2\).

Detailed Answer

a) We know that the bearing is \( 78^\circ \), we know that the angle \( \hat{LMB} = 12^\circ \), since we know Leipzig is directly east to Munich. Then:

\[\hat{BLM} = 180^\circ - 142^\circ - 12^\circ = 26^\circ\]

b) Let \( d \) be the distance from Munich to Berlin. We have all angles so it can be easily calculated with the sine rule:

\[\frac{d}{\sin(26)} = \frac{200}{\sin(12)}\]

\[ d = 421.689...\]

So the total distance traveled by Jack will be:

\[ 200 + 421.689... = 621.689...\]

Knowing that trains go with a constant speed we can calculate the time spent on the train:

\[ t = \frac{621.689...}{200} = 3.108... \ \text{hours} \]

Since one more hour was spent to change trains, the final time is 4.11 hours.

d) The area \( A \) of triangle \( MBL \) can be found using:

\[A = \frac{1}{2} \cdot a \cdot b \cdot (\sin(c) \]

\[A = \frac{1}{2} \cdot 421.689... \cdot 200 \cdot \sin(142) \approx 26000 \text{ km}^2 \]

Detailed Answer

a) Using the cosine rule we have: \(x^2 = a^2 + b^2 - 2ab\cos(\gamma)\), so \(x = \sqrt{1.5^2 + 0.5^2 - 2 \times 1.5 \times 0.5 \times \cos(105°)} = 1.7\,\text{km}\)

b) Using the cosine rule: \(47.6^2 = 40^2 + 14^2 - 2 \times 40 \times 14 \times \cos(x)\). So, \(x = \cos^{-1}\left(\frac{-(47.6^2 - 40^2 - 14^2)}{2 \times 40 \times 14}\right) = 115°\). If we want the angle between 0 and 90°, this gives us 65°.

c) We added in blue on the diagram the angles that we can deduce ourselves.

The cosine rule gives us: \(b = \sqrt{550^2 + 300^2 - 2 \times 550 \times 300 \times \cos(135°)} = 790\,\text{km}\). For the bearing, we will use the following diagram, where we have highlighted in green more angles that can be deduced:

We start by finding the angle of the triangle at point C, written \(\gamma\), thanks to the sine rule:

\[\frac{c}{\sin(\gamma)} = \frac{b}{\sin(135°)} \implies \gamma = \sin^{-1}\left(\frac{c \times \sin(135°)}{b}\right) = 30°\]

By reinjecting the exact value of \(b\) we found before. Finally, the bearing is \(360°\) minus the right side of \(\gamma\), so: \( \text{bearing} = 360° − (\gamma − 20°) = 350°\).

Detailed Answer

a)

b) The size of the side of the triangle is \(100 \times \tan(30°) = 58\,\text{m}\). Adding Jerry’s height, we find \(d = 59\,\text{m}\).

c)

d) \(𝛽 = \tan^{-1}\left(\frac{20}{50}\right) = 22°\)

e)

f) This is a right-angled triangle, and we can find the length of the hypotenuse using Pythagoras, so: \( \text{distance} = \sqrt{(59 + 20)^2 + 50^2} = 94\,\text{m}\)

Detailed Answer

a)\(x_I = \frac{-1+5}{2}=2\) and \(y_I = \frac{-1+2}{2}=0.5\)

b) The equation of line (AB) has for gradient \(a = \frac{-2+1}{5+1}=\frac{1}{2}\) and \(y_I = \frac{-1+2}{2}=0.5\). The equation of a perpendicular line must have a gradient such that \(a' * a = -1 \implies a' = -2\)

We find the intercept by plugging the coordinates of I into the equation: \(y=-2x + b\)

\[ 0.5 = -2 * 2 + b \implies b = 4.5\]

Thus, the equation is \(y=-2x + 4.5\)

c) M(0,4.5)

d) \([BI] = \sqrt{(5-2)^2 + (2-0.5)^2} = \sqrt{11.25} = 3.35\)

\([IM] = \sqrt{(2-0)^2 + (0.5-4.5)^2} = \sqrt{20} = 4.47\)

e) M is on the bisector of [AB], so this is a right triangle, \(area_{MIB} = \frac{1}{2}\sqrt{11.25} * \sqrt{20} = 7.5\)

f) By symmetry around the bisector, this is twice the area of the previous triangle, so 𝑎𝑟𝑒𝑎_MIB = 15.

Detailed Answer

a) \[\hat{AQB} = 180^\circ - \hat{QBC} = 180^\circ - 120^\circ = 60^\circ\]

b) To find the length of \( QC \), we use the fact that \( Q \) is the midpoint of \( AC \), so \( QC = 6 \text{cm} \)

c) To find \( \hat{ABQ} \), we use the sine rule in \( \triangle ABQ \):
\[\frac{AB}{\sin(\hat{AQB})} = \frac{AQ}{\sin(\hat{ABQ})}\]
Substituting the known values:
\[\frac{7}{\sin(60^\circ)} = \frac{6}{\sin(\hat{ABQ})}\]
\[\sin(\hat{ABQ}) = \frac{6 \cdot \sin(60^\circ)}{7}\]
\[\hat{ABQ} \approx 47.9^\circ\]

d) \[\hat{BAQ} = 180^\circ - 60^\circ - 47.9^\circ \approx 72.1^\circ\]

Detailed Answer

a) To find the length of \( AB \), we use the cosine rule:
\[ AB^2 = BC^2 + AC^2 - 2 \cdot BC \cdot AC \cdot \cos(\hat{ACB}) \] \[ AB^2 = 320^2 + 410^2 - 2 \cdot 320 \cdot 410 \cdot \cos(61^\circ) \] \[ AB \approx 389 \text{m} \]

b) To find the cost of filling the car park with cement, we first calculate the area of \( \triangle ABC \) using the formula: \[\text{Area} = \frac{1}{2} \cdot BC \cdot AC \cdot \sin(\hat{ACB}) \] \[\text{Area} = \frac{1}{2} \cdot 320 \cdot 410 \cdot \sin(61^\circ) \] \[\text{Area} = 57375.05... \text{ m}^2\]
The cost to fill the car park with cement is:
\[\text{Cost} = 57375.05... \cdot 0.3 \approx 17213 \text{ USD}\]
Close

Detailed Answer

a) To find the length of \(AC\), we apply the cosine rule:
\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\hat{ABC}) \] \[ AC^2 = 8^2 + 6^2 - 2 \cdot 8 \cdot 6 \cdot \cos(92^\circ) \] \[ AC = 10.16... \approx 10.2 \text{m} \]

b) \[ \text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot \sin(92^\circ) \approx 24.0 \text{m}^2 \]

c) For this part we should implement the sine rule with the answer from part (a): \[ \frac{\sin(110)}{10.16...} = \frac{\sin(\hat{ACD})}{5} \] \[ \hat{ACD} = 27.52... \approx 27.5^\circ \]

d) The area of quadrilateral \( ABCD \) can be found by summing the areas of \( \triangle ABC \) and \( \triangle ACD \). The area of \( \triangle ABC \) is approximately \( 24.0 \, \text{m}^2 \). To calculate the area of \( \triangle ACD \) we need some additional information, for example the size of the angle \( \hat{DAC} \) would be useful to the apply the formula for the area of the triangle. \[ \hat{DAC} = 180 - 110 - 27.52... = 42.47... \approx 42.5^\circ \] \[ \text{AreaADC} = \frac{1}{2} \cdot 5 \cdot 10.16... \cdot \sin(42.47...^\circ) = 17.16... \approx 17.2 \text{m}^2 \] \[ \text{AreaABCD} = 17.16... + 23.99... \approx 41.1 \text{m}^2 \]

Detailed Answer

a) \(area = 𝜋 * 10^2 * \frac{70°}{360°} = 61m^2\)

b) \(arc = \frac{70°}{360°} * 2𝜋 * 10 = 12m\)

c)

d) We start by calculating the area of roses:

\[ roses = \frac{70°}{360°} * 𝜋 * 2^2 = 2.4m^2\]

Then to find the area of tulips, we calculate as if it were a full sector of radius (5+2), then subtract the area of roses. Be careful not to use 5 as the radius, as this would be the area of a full sector of radius 5.

\[ tulips = \frac{70°}{360°} * 𝜋 * (5+2)^2 - roses = 27m^2\]

Finally, for the daisies, we subtract the two previous areas to the total garden area. We find:

\[ daisies = 61m^2 - 27m^2 - 2.4m^2 = 31m^2\]

e) The path between tulips and roses is the arc of a circle of radius 2 meters.

\[ path_1 = \frac{70°}{360°} * 2𝜋 * 2 = 2.4m\]

The path between tulips and roses is the arc of a circle of radius (5+2) meters.

\[ path_2 = \frac{70°}{360°} * 2𝜋 * 7 = 8.6m\]

Detailed Answer

a) To find \( BE \), we use the distance formula in 3D:

\[ BE = \sqrt{(1 - (-3))^2 + (8 - 1)^2 + (0 - 8)^2} = \sqrt{4^2 + 7^2 + (-8)^2} = \sqrt{129} \]

b) We are given that the angle \( \hat{ABE} = 68^\circ \). We know that both angles from the base have to be the same, so we also know that \( \hat{AEB} = 44^\circ \). Thus, we can use the sine rule:

\[ \frac{\sin(68)}{\sqrt{129}} = \frac{\sin(44)}{AB} \]

\[ AB = 8.51 \]

c) \( \text{Area of base} = AB^2 = 72.4 \text{cm}^2 \)

d) The volume of the cylinder is equal to the volume of the pyramid, so we use the volume formula for the cylinder:

\[ V = \pi r^2 h = 720 \Rightarrow h = \frac{720}{\pi r^2} \]

The surface area of the cylinder is given by:

\[ S = 2\pi r h + 2\pi r^2 \]

So, by plugging the expression for \( h \):

\[ S = \frac{1440}{r} + 2\pi r^2 \]

e) This can be easily done by plugging the function obtained in part (d) in the GDC:

Thus, \( r = 4.86 \text{cm} \).

Detailed Answer

a) This funciton has an amplitude of 28, so:

\[ \text{min} = 30 - 28 = 2 \text{m}\]

\[ \text{max} = 30 + 28 = 58 \text{m}\]

b) To find the altitude after 2 minutes, substitute \( t = 2 \) into the height function:

\[ h(2) = 28 \cos \left( \frac{\pi}{4}(2 - 3) \right) + 30 = 28 \cos \left( -\frac{\pi}{4} \right) + 30 \approx 49.8 \text{m} \]

c) To determine when the passenger reaches maximum altitude, we can set the cosine term to 1:

\[ \frac{\pi}{4}(t - 3) = 0 \Rightarrow t - 3 = 0 \Rightarrow t = 3 \, \text{minutes} \]

The function can also be plugged into the GDC to find the maximum.

d) We can plot the function and the straight horizontal line of \( y = 40 \) into the GDC and find the first interesections:

We do the same with the second intersection to find that the passenger is above the height of 40m between \( t=1.46 \) and \( t=4.54 \), so:

\[ 4.54 - 1.46 = 3.08 \ \text{minutes}\]

Detailed Answer

a) \( h(12) = 5 \cos (4\pi) + 6 = 5 \cdot 1 + 6 = 11 \, \text{meters} \)

b) We are looking to find the period:

\[ \frac{\pi}{3} \cdot T = 2\pi \implies T=6 \, \text{hours} \]

c) The maximum height is when \( \cos \left( \frac{\pi}{3} t + \frac{\pi}{3} \right) = 1 \):

\[ h_{max} = 5 \cdot 1 + 6 = 11 \, \text{meters} \]

The minimum height is when \( \cos \left( \frac{\pi}{3} t + \frac{\pi}{3} \right) = -1 \):

\[ h_{min} = 5 \cdot (-1) + 6 = 1 \, \text{meter} \]

d) We can use the GDC and graph this function to find the values that satisfy the criterion between \( t=10 \) and \( t=18 \):

So, we can see that the high tide will be at 11AM, and using the period we found in part (b) we also know that it will happen at 5PM. In the same way we can find the low tide which will happen at 2PM (when \( x=14 \)).

e) The height must be less than or equal to 7 meters for swimming, so we can plot it in the GDC:

So, after finding the two interesects we get:

\[ 15.69... - 12.307... \approx 3.38 = 3 \, \text{hours} \, 23 \, \text{minutes} \]