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 ∠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 4ABC by multiplying the length of the altitude (the height) by
the corresponding base and dividing by 2. Therefore,
AQ × BC
BR × AC
CP × AB
=
=
2
2
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 4ARB, sin A = BR
=
AB
24
56
together gives c = b , or 24b = 56c.
24
.
c
In 4AP C, sin A =
CP
AC
AQ
In right-angled 4BAQ, sin B = AB
=
21
56
together gives c = a , or 21a = 56c.
21
.
c
In 4BP C, sin B =
CP
BC
Combining these gives 21a = 24b = 56c.
=
=
56
.
b
56
.
a
Putting these
Putting these
We now will continue on with the rest of the solution.
21
21
From 21a = 24b we obtain b = 24
a = 78 a, and from 21a = 56c we obtain c = 56
a = 38 a.
The ratio of the sides in 4ABC is therefore a : b : c = a : 78 a : 38 a = 8 : 7 : 3. Let BC = 8x,
AC = 7x, and AB = 3x, where x > 0.
Using the cosine law,
AC 2 = AB 2 + BC 2 − 2(AB)(BC) cos(∠ABC)
(7x)2 = (3x)2 + (8x)2 − 2(3x)(8x) cos(∠ABC)
49x2 = 9x2 + 64x2 − 48x2 cos(∠ABC)
Since x > 0, we know x2 6= 0. So dividing by x2 ,
49 = 73 − 48 cos(∠ABC)
Rearranging,
48 cos(∠ABC) = 24
1
cos(∠ABC) =
2
Therefore, ∠ABC = 60◦ .
In right 4BP C,
CP
= sin 60◦
BC
CP
BC =
sin 60◦
56
BC = √
3
2
√
112
3
BC = √ × √
3
3
√
112 3
BC =
3
However, BC = 8x. Therefore,
√
112 3
8x =
3
√
14 3
x=
3
√
3x = 14 3
√
98 3
7x =
3
√
◦
Therefore, ∠ABC
√ = 60 , and the side√lengths of 4ABC are AB = 3x = 14 3 cm,
98 3
112 3
AC = 7x =
cm, and BC =
cm.
3
3