Solve Applications Involving Angles

When investigating the geometrical relationships in an arrangement of lines and angles, we make use of established results about, for example, angles on a line, the angle sum of a triangle, angles formed with parallel lines, similar triangles and symmetries.

There are various ways of showing that the angle sum of a triangle is $180^\circ$180°. Here is one method.

The triangle is formed by the rays, drawn with arrow heads. Moving anticlockwise around the triangle, we see that each ray can be transformed into the next by an anticlockwise rotation about a vertex and a suitable translation.

Rotating a ray about each vertex in turn, brings it back to its starting position. It has undergone a full $360^\circ$360°rotation and in the process, it has turned through the three exterior angles, Thus, we see that

$360^\circ=\left(180^\circ-a\right)+\left(180^\circ-b\right)+\left(180^\circ-b\right)$360°=(180°−a)+(180°−b)+(180°−b)and therefore, $a+b+c=180^\circ$a+b+c=180°.

To illustrate the mathematical thinking that might be involved, we investigate the relationships to be found in a pentagon with its diagonals.

A simple pentagon, whose boundary does not cross over itself, can always be divided into three triangles. The nine angles in the triangles must be equivalent to the five angles of the pentagon. So, we see that the angle sum of a pentagon must be $3\times180^\circ=540^\circ$3×180°=540°. 

In the case of a regular pentagon - all sides equal, all angles equal - it must be that each angle is $540^\circ\div5=108^\circ$540°÷5=108°.

The following pentagon diagram was constructed by repeatedly copying and rotating a side through the angle $108^\circ$108° and then translating the sides to form the polygon.

The diagonal rays form a five-pointed star shape. We can investigate the size of the angles at the star points using a similar strategy to the one that was used above for the angle sum of a triangle. We also make use of the symmetry properties: the regular pentagon can be rotated about its centre five times through the same angle so that it coincides with its original position at each rotation, and there are five axes of symmetry such that the pentagon reflected in any one of them will look unchanged.

Starting with the downward pointing ray, we rotate it anticlockwise through the angle at each star point. After five such rotations, the ray is aligned with its original position but is now pointing in the opposite direction. The ray has turned through $180^\circ$180° made up of the five equal star-point angles. Thus, the angle at each star point must be $180\div5=36^\circ$180÷5=36°. 

We have established that the angle $\alpha=36^\circ$α=36°. By symmetry, we know that $\beta_1=\beta_2$β1​=β2​ and $\gamma_1=\gamma_2$γ1​=γ2​. Combining the facts that $\beta_1+\alpha+\beta_2=108^\circ$β1​+α+β2​=108°, $\alpha=36^\circ$α=36° and $\beta_1=\beta_2$β1​=β2​, we see that $\beta=\frac{108-36}{2}=36^\circ$β=108−362​=36°.

Again by symmetry, we have $\gamma_1=\gamma_2$γ1​=γ2​. So, $\gamma=\frac{180-108}{2}=36^\circ$γ=180−1082​=36°.

So, we have shown that all the acute angles formed at the vertices of a regular pentagon by its diagonals are $36^\circ$36° angles.

From the fact that alternate angles marked $\alpha$α in the above diagram are equal, we see that a diagonal and a side of the pentagon are parallel. Moreover, each of the five diagonals is parallel to a side.

Referring to the earlier diagram with all the diagonals drawn in the pentagon, we can see that there are two different triangle shapes formed and there are fifteen of each. That is, there are two sets, each of fifteen similar triangles. We can use equivalent side ratios from the similar triangles to deduce the side lengths.

In this diagram, we assume the pentagon has unit sides. We find the ratios $\frac{x}{1}=\frac{1}{x+1}$x1​=1x+1​. This equation, after rearrangement, is seen to be is quadratic in $x$x. 

$x^2+x-1=0$x2+x−1=0

It has the positive solution $x=\frac{\sqrt{5}-1}{2}\approx0.618$x=√5−12​≈0.618. This number occurs in other places in connection with the golden ratio.  It is the reciprocal of $\phi=\frac{\sqrt{5}+1}{2}\approx1.618$ϕ=√5+12​≈1.618. 

If a perpendicular bisector of a side is inserted, it must also bisect a vertex angle. 

From this, we see that $\sin18^\circ=\frac{\frac{1}{2}}{1+\frac{\sqrt{5}-1}{2}}=\frac{1}{1+\sqrt{5}}$sin18°=12​1+√5−12​​=11+√5​, which is another exact value of a trigonometric function.

 

 

 

 

 

 

Worked Examples

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Question 3