Question

In ΔABC, AD ⊥ BC, AB = 13 cm, BD = 5 cm, DC = 4 cm. Find the value of 

1. AD
2. tan x + cot y

Answer 1

(i) AD:

ΔABD,

AB² = AD² + BD²

13² = AD² + 5²

= AD² = 169 − 25

= 144 = AD

AD = 12cm

(ii) tan ⁡x + cot ⁡y:

In △ACD, the angle at A is x.

tan x = `(DC)/(AD)`

= `4/12`

= `1/3`

In right ΔACD, the angle at A is x.

tan x = `(DC)/(AD)`

= `4/12`

= `1/3`

In the right △ABD angle at B is y.

tan y = `(AD)/(BD)` 

= `(AD)/(BD) =`

= `12/5`

= cot y = `5/12`

so,

= tan x + cot y    ...[Adding]

= `1/3 + 5/12`

= `4/12 + 5/12`

= `9/12`

= `3/4`