UPSEE 2019 Solved Paper

Given, In ΔABC
∠B=90°
BD is perpendicular on AC
AC=4BD
In ΔABD
⇒tan‌A=

BD

AD

........(i)
AD=BD‌cot‌A
In ΔBDC
⇒ tan‌C=

BD

CD

CD=BD‌cot‌C ........(ii)
From Eqs. (i) and (ii)
AD+CD=BD(cot‌A+cot‌C)
AD+CD=BD(cot‌A+cot‌C)
AC=BD(cot‌A+cot(

π

2

−A)) [∵A+C=

π

2

]
4BD=BD(cot‌A+tan‌A)
4=

cos‌A

sin‌A

+

sin‌A

cos‌A

⇒ 4=

1

sin‌A‌cos‌A

⇒ 2=

1

2‌sin‌A‌cos‌A

⇒ sin‌2‌A=

1

2

⇒ 2A=30°,150°
⇒ A=15°,75°
⇒ Hence, C=75°,15°
∴ Smallest angle is 15° or,

π

12