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) \(c = \frac{9.2}{tan(49°)}\)
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 \(b = 12\).
a) We use the formula for the area of a triangle: \(\frac{1}{2} \times 2 \times 2 \sin(60°) = 1.72\,m^2\)
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 \(4.87\,m^2\).
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: \(x = \sqrt{10^2 - 3.5^2} = 9.4km\)
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: \( \text{bearing} = 360° − (\gamma − 20°) = 350°\).
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: \( \text{distance} = \sqrt{(59 + 20)^2 + 50^2} = 94\,\text{m}\)
Closea)\(x_I = \frac{-1+5}{2}=2\) 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 𝑎𝑟𝑒𝑎MIB = 15.
Closea) \(area = 𝜋 * 10^2 * \frac{70°}{360°} = 61m^2\)
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.
\[ path_2 = \frac{70°}{360°} * 2𝜋 * 7 = 8.6m\]
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