11.2 Compound angle identities and applications  (Page 3/3)

Problems in 3 dimensions

$D$ is the top of a tower of height $h$ . Its base is at $C$ . The triangle $ABC$ lies on the ground (a horizontal plane). If we have that $BC=b$ , $D\widehat{B}A=\alpha$ , $D\widehat{B}C=x$ and $D\widehat{C}B=\theta$ , show that

$$h=\frac{bsin\alpha sinx}{sin(x+\theta )}$$

1. Identify a strategy

   We have that the triangle $ABD$ is right-angled. Thus we can relate the height $h$ with the angle $\alpha$ and either the length $BA$ or $BD$ (using sines or cosines). But we have two angles and a length for $▵BCD$ , and thus can work out all the remaining lengths and angles of this triangle. We can thus work out $BD$ .

2. Execute the strategy

   We have that

   $$\begin{array}{ccc}\hfill \frac{h}{BD}& =& sin\alpha \hfill \\ \hfill \Rightarrow h& =& BDsin\alpha \hfill \end{array}$$

   Now we need $BD$ in terms of the given angles and length $b$ . Considering the triangle $BCD$ , we see that we can use the sine rule.

   $$\begin{array}{ccc}\hfill \frac{sin\theta }{BD}& =& \frac{sin\left(B\widehat{D}C\right)}{b}\hfill \\ \hfill \Rightarrow BD& =& \frac{bsin\theta }{sin\left(b\widehat{D}C\right)}\hfill \end{array}$$

   But $D\widehat{B}C={180}^{\circ }-\alpha -\theta$ , and

   $$\begin{array}{ccc}\hfill sin({180}^{\circ }-\alpha -\theta )& =& -sin(-\alpha -\theta )\hfill \\ & =& sin(\alpha +\theta )\hfill \end{array}$$

   So

   $$\begin{array}{ccc}\hfill BD& =& \frac{bsin\theta }{sin\left(D\widehat{B}C\right)}\hfill \\ & =& \frac{bsin\theta }{sin(\alpha +\theta )}\hfill \end{array}$$

1. The line $BC$ represents a tall tower, with $C$ at its foot. Its angle of elevation from $D$ is $\theta$ . We are also given that $BA=AD=x$ .

   

   1. Find the height of the tower $BC$ in terms of $x$ , $tan\theta$ and $cos2\alpha$ .
   2. Find $BC$ if we are given that $x=140m$ , $\alpha ={21}^{\circ }$ and $\theta =9^{\circ }$ .

Other geometries

Taxicab geometry

Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in which the usual metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the (absolute) differences of their coordinates.

Manhattan distance

The metric in taxi-cab geometry, is known as the Manhattan distance , between two points in an Euclidean space with fixed Cartesian coordinate system as the sum of the lengths of the projections of the line segment between the points onto the coordinate axes.

For example, the Manhattan distance between the point $P_1$ with coordinates $(x_1,y_1)$ and the point $P_2$ at $(x_2,y_2)$ is

The Manhattan distance changes if the coordinate system is rotated, but does not depend on the translation of the coordinate system or its reflection with respect to a coordinate axis.

Manhattan distance is also known as city block distance or taxi-cab distance. It is given these names because it is the shortest distance a car would drive in a city laid out in square blocks.

Taxicab geometry satisfies all of Euclid's axioms except for the side-angle-side axiom, as one can generate two triangles with two sides and the angle between them the same and have them not be congruent. In particular, the parallel postulate holds.

A circle in taxicab geometry consists of those points that are a fixed Manhattan distance from the center. These circles are squares whose sides make a ${45}^{\circ }$ angle with the coordinate axes.

The great-circle distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere's interior). Because spherical geometry is rather different from ordinary Euclidean geometry, the equations for distance take on a different form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere). The shape of the Earth more closely resembles a flattened spheroid with extreme values for the radius of curvature, or arcradius, of 6335.437 km at the equator (vertically) and 6399.592 km at the poles, and having an average great-circle radius of 6372.795 km.

Summary of the trigonomertic rules and identities

End of chapter exercises

Do the following without using a calculator.

1. Suppose $cos\theta =0.7$ . Find $cos2\theta$ and $cos4\theta$ .
2. If $sin\theta =\frac{4}{7}$ , again find $cos2\theta$ and $cos4\theta$ .
3. Work out the following:
   1. $cos{15}^{\circ }$
   2. $cos{75}^{\circ }$
   3. $tan{105}^{\circ }$
   4. $cos{15}^{\circ }$
   5. $cos{3}^{\circ }cos{42}^{\circ }-sin{3}^{\circ }sin{42}^{\circ }$
   6. $1-2{sin}^2(22.5^{\circ })$
4. Solve the following equations:
   1. $cos3\theta ·cos\theta -sin3\theta ·sin\theta =-\frac{1}{2}$
   2. $3sin\theta =2{cos}^2\theta$
5. Prove the following identities
   1. ${sin}^3\theta =\frac{3sin\theta -sin3\theta }{4}$
   2. ${cos}^2\alpha (1-{tan}^2\alpha )=cos2\alpha$
   3. $4sin\theta ·cos\theta ·cos2\theta =sin4\theta$
   4. $4{cos}^{3}x-3cosx=cos3x$
   5. $tany=\frac{sin2y}{cos2y+1}$
6. (Challenge question!) If $a+b+c={180}^{\circ }$ , prove that
   $${sin}^{3}a+{sin}^{3}b+{sin}^{3}c=3cos(a/2)cos(b/2)cos(c/2)+cos(3a/2)cos(3b/2)cos(3c/2)$$