19 (a) Write down the value of sin 45° - OCR - GCSE Maths - Question 19 - 2019 - Paper 1

To find the length of BC, use triangle ADB:

1. In triangle ADB,
   • AD = 10/√6 mm
   • Angle BAD = 30°
   • Therefore, using the definition of tangent: [ \tan(30°) = \frac{AB}{AD} ] Substitute ( \tan(30°) = \frac{1}{\sqrt{3}} ): [ \frac{1}{\sqrt{3}} = \frac{AB}{10/\sqrt{6}} ] Rearranging gives: [ AB = \frac{10/\sqrt{6}}{\sqrt{3}} = \frac{10}{\sqrt{18}} = \frac{10}{3\sqrt{2}} ] Thus, ( AB = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3} ).
2. Since BC = CD, we can find BC using triangle BCD:
   • Angle BDC = 90°
   • BD = AB = ( \frac{5\sqrt{2}}{3} )
   Applying Pythagorean theorem: [ CD^2 + BC^2 = BD^2 ] Replace CD with BC: [ BC^2 + BC^2 = \left(\frac{5\sqrt{2}}{3}\right)^2 ] [ 2BC^2 = \frac{50}{9} ] [ BC^2 = \frac{25}{9} ] [ BC = \frac{5}{3} \text{ mm} ]