Question

Find the length of AD.
Given : angle ABC = 60°,
angle DBC =
= 45°
and BC =40 cm​

Answer 1

● Given :-

• ∠ABC = 60°
• ∠DBC = 45°
• Line BC = 40cm

● We need to find :-

• Length of line AD = ?

● Solution :-

Here, ∆ACB is a right triangle with right angle at C & ∆ACB have a right angle traingle in it ∆DCB

Firstly finding the height ( h ) of ∆ACB using trigonometric identity,

❑ tanθ = opposite/adjacent

Here ,

• Opposite side of ∠ABC = Height = AC
• Adjacent side of ∠ABC = Base = BC = 40cm

Substituting θ = 60° = ∠ABC & known values

→ tan60° = AC/BC

→ √3 = AC/40cm

[ ∵ tan60° = √3 & base = BC = 40cm ]

→ Height ( AC ) = 40√3 cm

Now , finding Height ( DC ) of ∆DCB using trigonometric identity tanθ

Here ,

• Opposite side of ∠DBC = Height = DC
• Adjacent side of ∠DBC = Base = BC = 40cm

Substituting θ = 45° = ∠DBC & known values

⇒ tan45° = DC/BC

⇒ 1 = DC/40

[ ∵ tan45° = 1 & base = BC = 40cm ]

⇒ DC = 40cm

_________________________

Now , observing the figure

AD = AC - DC

Substituting

• AC = 40√3
• DC = 40

➵ AD = 40√3 - 40 = 40( √3 - 1 )

Hence , AD = 40( √3 - 1 ) cm

Answer 2

Answer:

• BC = 40 cm
• ∠ DBC = 45°
• ∠ ABC = 60°

• In Triangle DBC :

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(45)=\dfrac{DC}{BC}\\\\\\:\implies\sf 1 =\dfrac{DC}{40}\\\\\\:\implies\sf 40 = DC\qquad...eq\:(I)$

⠀

• In Triangle ABC :

$:\implies\sf \tan(\theta)=\dfrac{Perpendicular}{Base}\\\\\\:\implies\sf \tan(60)=\dfrac{AC}{BC}\\\\\\:\implies\sf \sqrt{3} =\dfrac{AC}{40}\\\\\\:\implies\sf 40 \sqrt{3} = AC\\\\\\:\implies\sf40 \sqrt{3} = AD+DC\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting value of DC from eq. ( I )}}\\\\:\implies\sf 40 \sqrt{3} =AD+40\\\\\\:\implies\sf 40 \sqrt{3} - 40 = AD\\\\\\:\implies\sf 40( \sqrt{3} - 1) = AD\\\\\\:\implies\underline{\boxed{\sf AD = 40( \sqrt{3} - 1)\:cm}}$