Using the measurements given in the figure alongside,

(a) Find the values of:

(i) sin Φ

(ii) tan θ.

(b) Write an expression for AD in terms of θ.

(a) In right-angled ∆BCD

Using pythagoras theorem we get :

⇒ BD² = BC² + CD²

⇒ CD² = BD² - BC²

⇒ CD² = (13)² - (12)²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$ = 5.

(i) By formula,

sin Φ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

In right-angled ∆BCD

sin Φ = $\dfrac{CD}{BD} = \dfrac{5}{13}$.

Hence, sin Φ = $\dfrac{5}{13}$.

(ii) Draw DE perpendicular to AB.

From figure,

ED = BC = 12

In right-angled ∆BED

Using pythagoras theorem we get :

⇒ BD² = ED² + EB²

⇒ 13² = 12² + EB²

⇒ EB² = (13)² - (12)²

⇒ EB² = 169 - 144

⇒ EB² = 25

⇒ EB = $\sqrt{25}$ = 5.

From figure,

AE = AB - EB = 14 - 5 = 9.

By formula,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{ED}{AE} = \dfrac{12}{9} = \dfrac{4}{3}$.

Hence, tan θ = $\dfrac{4}{3}$.

(b) In right-angled ∆AED

$\phantom{\Rightarrow} \text{sin θ} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{sin θ} = \dfrac{ED}{AD} \\[1em] \Rightarrow \text{sin θ} = \dfrac{12}{AD} \\[1em] \Rightarrow AD = \dfrac{12}{\text{sin θ}}. \\[1em] \phantom{\Rightarrow} \text{cos θ} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{cos θ} = \dfrac{AE}{AD} \\[1em] \Rightarrow \text{cos θ} = \dfrac{9}{AD} \\[1em] \Rightarrow AD = \dfrac{9}{\text{cos θ}}.$

Hence, AD = $\dfrac{9}{\text{cos θ}} \text{ or } \dfrac{12}{\text{sin θ}}$.