The diagram shows triangle ABC with D on AC and E on AB - OCR - GCSE Maths - Question 14 - 2017 - Paper 1

Since DE is a straight line, we can find angle DAE using:

$DAE = 180^{\circ} - A - 72^{\circ}$

Substituting the value of angle A:

$DAE = 180^{\circ} - 36.3^{\circ} - 72^{\circ} = 71.7^{\circ}$

Now we can apply the sine rule in triangle ACD:

$\frac{CD}{\sin A} = \frac{AC}{\sin DAE}$

First, we need to find the length AC:

Using the cosine rule again:

$AC^2 = AD^2 + DE^2 - 2 \cdot AD \cdot DE \cdot \cos A$

Substituting for AC:

• Using the earlier found values, we get:

$AC^2 = 28^2 + 22^2 - 2 \cdot 28 \cdot 22 \cdot \cos(36.3^{\circ})$

Calculating these terms will yield the necessary lengths to find CD:

Rearranging the sine rule for CD then gives:

$CD = \frac{AC \cdot \sin A}{\sin DAE}$