Pythagoras` Theorem and the Trigonometric ratios (sin, cos, tan) require right-angled triangles. Both the theorem and the trigonometric ratios can be used on more general triangles that do not contain a right angle.
Triangles can be divided into two smaller triangles:
In this instance, the vertical height is the same for both smaller triangles. This vertical height can also be used to work out the area of the triangles.
If angles are given, the sin rule or the cosine rule are more likely to be used to solve problems involving triangles.
A triangle ABD is shown below. A vertical line dropped from A intersects the line BD at C. The ratio of the lengths BC:CD is 2:7. What is the distance BD?
The distance AC is the same for both triangles ABC and ACD
Let the distance BC be 2y and CD be 7y
AC same for both | `sqrt(8^2 - (2y)^2)` | `= sqrt(12^2 - (7y)^2)` |
`8^2 - 4y^2` | `= 12^2 - 49y^2` | |
`45y^2` | `= 80` | |
`y^2` | `= 1.7778` | |
`y` | `= 1.3333` | |
Length of BD | `= 2y + 7y` | |
substitute | `= 2(1.3333) + 7(1.333)` | |
`= 12` |
Answer: 12cm
What is the area of the triangle ABC, shown below? Give the answer to 1 decimal place.
Draw a vertical from A. Let this be length x
use SOHCAHTOA | `sin` | `=frac(text(opposite))(text(hypotenuse))` |
substitute | `sin (60)` | `=frac(x)(8)` |
`6.928` | `= x` | |
For a triangle | `text(Area)` | `= frac(1)(2) xx bh` |
Substitute | `= frac(1)(2) xx 6 xx 6.928` | |
` =20.785` | ||
To 1dp | ` =20.8` |
Answer: 20.8 cm2
See also Cosine Rule and Area of Any Triangle