Using the measurements given in the following figure :

(i) Find the value of sin Φ and tan θ.

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

(i) In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ 13² = 12² + CD²

⇒ 169 = 144 + CD²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$

⇒ CD = 5

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

$= \dfrac{CD}{BD} = \dfrac{5}{13}$

Draw a line parallel to BC from point D such that it meets AB at point E. This line DE will be ⊥ to AB.

From figure,

DE = BC = 12

In Δ BED,

⇒ BD² = BE² + DE² (∵ BD is hypotenuse)

⇒ 13² = BE² + 12²

⇒ 169 = BE² + 144

⇒ BE² = 169 - 144

⇒ BE² = 25

⇒ BE = $\sqrt{25}$

⇒ BE = 5

And, AE = AB - BE = 14 - 5 = 9

tan θ = $\dfrac{Base}{Perpendicular}$

$= \dfrac{DE}{AE} = \dfrac{12}{9} = \dfrac{4}{3}$

Hence, sin Φ = $\dfrac{5}{13}$ and tan θ = $\dfrac{4}{3}$

(ii) sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

sin θ = $\dfrac{DE}{AD} = \dfrac{12}{AD}$

AD = $\dfrac{12}{\text{sin θ}}$ = 12 cosec θ

cos θ = $\dfrac{Base}{Hypotenuse}$

cos θ = $\dfrac{AE}{AD} = \dfrac{9}{AD}$

AD = $\dfrac{9}{\text{cos θ}}$ = 9 sec θ

Hence, AD = 12 cosec θ or 9 sec θ.