(a)

Method I

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan 30° = $\frac{\mathrm{|BD|}}{\mathrm{50\; +\; d}}$

|BD| = tan 30° x (50 + d)

|BD| = 28.87 + 0.58d ----- eqn 1

tan 70° = $\frac{\mathrm{|BD|}}{\mathrm{d}}$

|BD| = tan 70° x d

|BD| = 2.74d ---- eqn 2

--- eqn 1 = --- eqn 2

2.74d = 28.87 + 0.58d

2.74d - 0.58d = 28.87

2.16d = 28.87

Divide both sides by 2.16

d = $\frac{28.87}{2.16}$ = 13.4 cm (one decimal place)

Method II

∠ ACB + ∠ BCD = 180°(Sum of angles on a straight line)

∠ ACB + 70° = 180°

∠ ACB = 180° - 70°

∠ ACB = 110°

∠ BAC + ∠ ACB + ∠ ABC = 180°(Sum of angles in a triangle)

30° + 110° + ∠ ABC = 180°

140° + ∠ ABC = 180°

∠ ABC = 180° - 140°

∠ ABC = 40°

$\frac{\mathrm{sin\; 40°}}{50}$ = $\frac{\mathrm{sin\; 30°}}{\mathrm{|BC|}}$

Cross multiply

|BC| x sin 40° = 50 x sin 30°

Divide both sides by sin 40°

|BC| = $\frac{\mathrm{50\; x\; sin\; 30°}}{\mathrm{sin\; 40°}}$ = 38.89 cm

d is adjacent to ∠ BCD (70°) and |BC| is hypothenuse.

|BC| = 38.89 cm

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

cos 70° = $\frac{\mathrm{d}}{\mathrm{38.89\; cm}}$

Multiply both sides by the denominator 38.89 cm

d = 38.89 cm x cos 70°

d = 13.3 cm

(b)

Method I

|BD| = 2.74d ---- eqn 2

But d = 13.4 cm

|BD| = 2.74 x 13.4 cm

|BD| = 36.7 cm (one decimal place)

Method II

|BC| = 38.89 cm

d = 13.3 cm

|BD|² + 13.3² = 38.89²

|BD|² = 38.89² - 13.3²

|BD|² = 1335.54

Take square root of both sides.

|BD| = $\sqrt{\mathrm{1335.54}}$ = 36.5 cm