 # Problem of the Week Problem E and Solution Now I Know My ABCs

## Problem

In triangle $$ABC$$, point $$P$$ lies on $$AB$$, point $$Q$$ lies on $$BC$$, and point $$R$$ lies on $$AC$$ such that $$AQ$$, $$BR$$, and $$CP$$ are altitudes with lengths $$21$$ cm, $$24$$ cm, and $$56$$ cm, respectively. Determine the measure, in degrees, of $$\angle ABC$$, and the lengths, in centimetres, of $$AB$$, $$BC$$, and $$CA$$.

Note the diagram is not drawn to scale.

## Solution

Let $$BC=a$$, $$AC=b$$, and $$AB=c$$. We will present two methods for determining that $$21a=24b=56c$$, and then continue on with the rest of the solution.

• Method 1: Use Areas

We can find the area of $$\triangle ABC$$ by multiplying the length of the altitude (the height) by the corresponding base and dividing by 2. Therefore, $\frac{AQ\times BC}{2}=\frac{BR\times AC}{2}=\frac{CP\times AB}{2}$ Substituting $$AQ=21$$, $$BR=24$$, and $$CP=56$$, and multiplying through by $$2$$ gives us $$21a=24b=56c$$.

• Method 2: Use Trigonometry

In right-angled $$\triangle ARB$$, $$\sin A=\frac{BR}{AB}=\frac{24}{c}$$. In $$\triangle APC$$, $$\sin A=\frac{CP}{AC}=\frac{56}{b}$$. Putting these together gives $$\frac{24}{c}=\frac{56}{b}$$, or $$24b=56c$$.

In right-angled $$\triangle BAQ$$, $$\sin B=\frac{AQ}{AB}=\frac{21}{c}$$. In $$\triangle BPC$$, $$\sin B=\frac{CP}{BC}=\frac{56}{a}$$. Putting these together gives $$\frac{21}{c}=\frac{56}{a}$$, or $$21a=56c$$.

Combining these gives $$21a=24b=56c$$.

We now will continue on with the rest of the solution.

From $$21a=24b$$ we obtain $$b=\frac{21}{24}a=\frac{7}{8}a$$, and from $$21a=56c$$ we obtain $$c=\frac{21}{56}a=\frac{3}{8}a$$.
The ratio of the sides in $$\triangle ABC$$ is therefore $$a:b:c=a:\frac{7}{8}a:\frac{3}{8}a=8:7:3$$. Let $$BC=8x$$, $$AC=7x$$, and $$AB=3x$$, where $$x>0$$.

Using the cosine law, \begin{aligned} AC^2&=AB^2+BC^2-2(AB)(BC)\cos(\angle ABC)\\ (7x)^2&=(3x)^2+(8x)^2-2(3x)(8x)\cos(\angle ABC)\\ 49x^2&=9x^2+64x^2-48x^2\cos(\angle ABC)\end{aligned} Since $$x>0$$, we know $$x^2\neq 0$$. So dividing by $$x^2$$, $49=73-48\cos(\angle ABC)$ Rearranging, \begin{aligned} 48\cos(\angle ABC)&=24\\ \cos(\angle ABC)&=\frac{1}{2}\end{aligned} Therefore, $$\angle ABC=60^\circ$$.

In right $$\triangle BPC$$, \begin{aligned} \frac{CP}{BC}&=\sin 60^\circ\\ BC&=\frac{CP}{\sin 60^\circ}\\ BC&=\frac{56}{\frac{\sqrt{3}}{2}}\\ BC&=\frac{112}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ BC&=\frac{112\sqrt{3}}{3}\end{aligned} However, $$BC=8x$$. Therefore, \begin{aligned} 8x&=\frac{112\sqrt{3}}{3}\\ x&=\frac{14\sqrt{3}}{3}\\ 3x&=14\sqrt{3}\\ 7x&=\frac{98\sqrt{3}}{3}\end{aligned}

Therefore, $$\angle ABC=60^\circ$$, and the side lengths of $$\triangle ABC$$ are $$AB=3x=14\sqrt{3}$$ cm, $$AC=7x=\dfrac{98\sqrt{3}}{3}$$ cm, and $$BC=\dfrac{112\sqrt{3}}{3}$$ cm.