Question

If the perpendicular AD divides the base of the ∆ABC such that BD, CD and AD are in ratio 2 : 3 : 6, then angle A is equal to –

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (C)

$\frac{\pi}{4}$

Answer 2

From ∆ADB, tan θ₁ = = $\left( \frac { 2 } { 6 } \right)$

or θ₁ = tan^–1 $\left( \frac { 2 } { 6 } \right)$ = tan^–1 $\left( \frac { 1 } { 3 } \right)$ ……..(1)

and from ∆ADC, tan θ₂ = = $\frac { 3 } { 6 }$

or θ₂ = tan^–1 $\left( \frac { 3 } { 6 } \right)$ = tan^–1 $\left( \frac { 1 } { 2 } \right)$ ……..(2)

Q ∠A = θ₁ + θ₂

∠A = tan^–1 $\left( \frac { 1 } { 3 } \right)$ + tan^–1 $\left( \frac { 1 } { 2 } \right)$

= tan^–1 $\frac { ( 1 / 3 ) + ( 1 / 2 ) } { 1 - ( 1 / 3 ) ( 1 / 2 ) }$ = tan^–1 (1) = $\frac { \pi } { 4 }$.