Question

Please help!!! (Ignore the AD it’s actually supposed to be DC)

Answer 1

Answer:

BD = 22, DC = 11√3

Step-by-step explanation:

In triangle ABC, ∠B = 45°, ∠C = 90°. Hence:

∠A + ∠B + ∠C = 180° (sum of angles in a triangle)

∠A + 45 + 90 = 180

∠A + 135 = 180

∠A = 45°

Using sine rule to find BC:

$\frac{BC}{sinA}=\frac{AB}{sinC} \\\\\frac{BC}{sin45}=\frac{11\sqrt{2} }{sin90} \\\\BC=\frac{11\sqrt{2}*sin45 }{sin90} \\BC=11$

In triangle BCD, ∠D = 30°, ∠C = 90°. Hence:

∠D + ∠B + ∠C = 180° (sum of angles in a triangle)

∠B + 30 + 90 = 180

∠B + 120 = 180

∠B = 60°

Using sine rule to find BD:

$\frac{BD}{sinC}=\frac{BC}{sinD} \\\\\frac{BD}{sin90} =\frac{11}{sin30} \\\\BD=\frac{11*sin90}{sin30}\\\\BD=22$

Using sin rule to find DC:

$\frac{DC}{sinB}=\frac{BC}{sinD} \\\\\frac{DC}{sin60} =\frac{11}{sin30} \\\\DC=\frac{11*sin60}{sin30}\\\\DC=11\sqrt{3}$