GCSE(H),

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:

If angles are given, the sine rule or the cosine rule are more likely to be used to solve problems involving triangles.

1. 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?

Answer: 12cm

The distance AC is the same for both triangles ABC and ACD

Let the distance BC be 2*y* and CD be 7*y*

`AC = 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`

`BD = 2y + 7y = 9y = 12`

2. What is the area of the triangle ABC, shown below?

Answer: 20.8 cm^{2}

Draw a vertical from A. Let this be length *x*

sin = `frac(text(opposite))(text(hypotenuse))`

`sin 60 = frac(x)(8)`

`x = 6.928`

Area of a triangle = `frac(1)(2) xx bh`

Area = `frac(1)(2) xx 6 xx 6.928`

A= 20.785 = 20.8 to 1dp

(Or use the formula for the area of any triangle. This question derives the formula.)

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