Further problems 5

Exercise 1
  1. A surveying team are trying to find the height of a hill. They take a "sight" on the top of the hill and find that the angle of elevation is 23∘27′23^\circ 27^\prime They move a distance of on level ground directly away from the hill and take a second "sight". From this point, the angle of elevation is 19∘46′19^\circ 46^\prime. Find the height of the hill, correct to the nearest metre. Note: With 23∘27′23^\circ 27^\prime we mean the angle is 23∘23^\circ and 2727 minutes, where a degree has 6060 minutes. We can express this angle with a decimal number. For example, 23.5∘23.5^\circ should then mean 23∘23^\circ and 3030 minutes. So we have 30′∼0.51′∼0.53027′∼0.530⋅27=0.45\begin{array}{rll} 30^\prime &\sim& 0.5\\[3pt] 1^\prime &\sim& \frac{0.5}{30}\\[3pt] 27^\prime &\sim& \frac{0.5}{30}\cdot 27=0.45 \end{array} So 23∘27′23^\circ 27^\prime equal 23.45∘23.45^\circ.
  2. A triangle is formed by two straight lines f(x)=−x+6f(x)=-x+6 and g(x)=2xg(x)=2x and the x–axis. Determine the angles and the area of the triangle.
  3. A rectangle has side lengths a=77a = 77 and b=30b = 30. Find the angle of intersection between the two diagonals.
  4. An isosceles triangle with base c=12c = 12 has an angle of γ=120∘\gamma = 120^\circ at the top (vertex).
    1. Two points are placed onto the base cc such that the points divide the base into three equal parts, and two lines are drawn from these two points to the vertex. Determine the three angles at the vertex.
    2. The angle γ\gamma at the vertex is divided into three equal parts, forming three triangles. Find the bases of these triangles.
  5. Consider the two straight lines given by g(x)=53x+2g(x)=\frac{5}{3}x+2 and h(x)=−0.5x−3h(x)=-0.5x-3. Determine the angle of intersection between
    1. gg and the xx-axis.
    2. gg and hh.
  6. Consider a diagonal of a cube. Determine the angle between this diagonal and
    1. the side of the cube
    2. the other diagonal
  7. Measured along the surface of the Earth, the distance between Hamburg and München is 610km610km. What is the maximal depth of a straight tunnel connecting the two town? Assume Earth is a perfect sphere with radius 6370km6370km.
  8. Two buildings, a house and a tower (see figure below). From a window in the house (height a=7ma=7m is the angle of elevation to the top of the tower β=17∘\beta=17^\circ. The angle of depression from the window to the base of the tower is α=11∘\alpha=11^\circ. Determine the distance ee to the tower and the height hh of the tower.
  9. Simplify:

    1. 1−cos⁡(α)⋅1+cos⁡(α)\sqrt{1-\cos(\alpha)}\cdot \sqrt{1+\cos(\alpha)}

    2. sin⁡(α)3+sin⁡(α)⋅cos⁡(α)2\sin(\alpha)^3+\sin(\alpha)\cdot \cos(\alpha)^2

    3. 1cos⁡(α)⋅tan⁡(α)\frac{1}{\cos(\alpha)\cdot\tan(\alpha)}

Solution
  1. 523.73m523.73 m
  2. α=tan⁡−1(2)=63.43...∘\alpha=\tan^{-1}(2)=63.43...^\circ, β=45∘\beta=45^\circ, γ=180∘−α−β=71.57...∘\gamma=180^\circ-\alpha-\beta=71.57...^\circ. Area A=6⋅4/2=12A=6\cdot 4/2=12.
  3. 42.57∘42.57^\circ
  4. We have
    1. 30∘,60∘,30∘30^\circ, 60^\circ, 30^\circ
    2. 4.739,2.522,4.7394.739, 2.522, 4.739
  5. It is
    1. 59.03∘59.03^\circ
    2. 85.60∘85.60^\circ
  6. It is
    1. 54.7∘54.7^\circ
    2. 70.5∘70.5^\circ
  7. ≈7.3km\approx 7.3 km
  8. e=36.01,h=18.01e = 36.01, h = 18.01
  9. It is
    1. sin⁡(α)\sin(\alpha)
    2. sin⁡(α)\sin(\alpha)
    3. 1sin⁡(α)\frac{1}{\sin(\alpha)}