Further problems 5

Exercise 1: Height of a hill

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.

Solution

523.73  m\qty{523.73}{m}

Exercise 2: Triangle from linear functions

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.

Solution

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.

Exercise 3: Rectangle diagonal intersection

A rectangle has side lengths a=77a = 77 and b=30b = 30. Find the angle of intersection between the two diagonals.

Solution

42.57∘42.57^\circ

Exercise 4: Isosceles triangle vertex division

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.

Solution

We have

  1. 30∘,60∘,30∘30^\circ, 60^\circ, 30^\circ
  2. 4.739,2.522,4.7394.739, 2.522, 4.739
Exercise 5: Intersection of straight lines

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.

Solution

It is

  1. 59.03∘59.03^\circ
  2. 85.60∘85.60^\circ
Exercise 6: Angles in a cube

Consider a diagonal of a cube. Determine the angle between this diagonal and

  1. the side of the cube

  2. the other diagonal

Solution

It is

  1. 54.7∘54.7^\circ
  2. 70.5∘70.5^\circ
Exercise 7: Earth curvature and tunnel depth

Measured along the surface of the Earth, the distance between Hamburg and München is 610  km\qty{610}{km}. What is the maximal depth of a straight tunnel connecting the two cities? Assume Earth is a perfect sphere with radius 6370  km\qty{6370}{km}.

Solution

≈7.3  km\approx \qty{7.3}{km}

Exercise 8: House and tower height

Two buildings, a house and a tower (see figure below). From a window in the house (height a=7  ma=\qty{7}{m} 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.

Solution

e=36.01,h=18.01e = 36.01, h = 18.01

Exercise 9: Trigonometric simplification

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

It is

  1. sin⁡(α)\sin(\alpha)
  2. sin⁡(α)\sin(\alpha)
  3. 1sin⁡(α)\frac{1}{\sin(\alpha)}