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) With Pythagoras, the distance to the starting point is: \(x = \sqrt{100^2 + 85^2} = 130km\). The whale is 130 km away.

b) With Pythagoras, the last line of the trail is equal to \(x = \sqrt{10^2 + 16^2} = 19km\). So in total, the trail is 45km long.

c) With Pythagoras: \(x = \sqrt{10^2 - 3.5^2} = 9.4km\)

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.

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