# Right Angle Triangles

```Right Angle Triangles
Trigonometry
Metric relations
BA&sup2; = BH X BC
A
CA&sup2; = CH X CB
AH&sup2; = HB X HC
AB X AC = AH X BC
C
B
H
Example: from page 173 in your
workbook, question 1 b)
A
Notice we are given the measurement of BC
which is 9.
You can solve for z by using
Pythagoras.
You can then solve for AH using
AB X AC = AH X BC
Then you can solve for x by
using BA&sup2; = BH X BC
Once you have x you can
calculate 9-x to find y.
z = 7.2
y = 5.76
x = 3.24
h = 4.32
z
5.4
h
B
x
y
C
H
9
Metric Relations (cont’d)
To solve triangles using metric relations, you may have to use a system of
equations. Example on page 173 # 1 a)
B
x
1.5
H
y
1.2
A
z
C
When looking at this question, we need to establish our “knowns” and
“unknowns”. So far, we have measurements for AH and AB. By using our metric
relations we can solve for x, y, and z. With only AB and AH, we do not have
enough information to use only one formula. We have to use our “substitution
method” to solve.
How?
Take the 3rd formula and the first formula
AH&sup2; = HB X HC
BA&sup2; = BH X BC
1.2&sup2; = xy
1.5&sup2; = x (x + y)
1.44 = xy
2.25 = x&sup2; + xy
Isolate y
y = 1.44 / x
substitute for y
2.25 = x&sup2;+x(1.44/x)
2.25 = x&sup2; + 1.44
2.25 – 1.44 = x&sup2;
x = 0.9
Now that you have x you can solve for y by using the first formula:
BA&sup2; = BH X BC
2.25 = xy
2.25 = (0.9) (y)
y = 1.6
And finally to solve for z we can use Pythagoras for the bigger triangle or the
fourth formula since we have 3 out of 4 variables:
AB X AC = AH X BC
(1.5)(z) = (1.2)(2.5)
z=2
```