Question

If in a abc, bc = 5, ca = 4, ab = 3 and d, e are the points on bc such that bd = de = ec, then

Answer 1

Given :  in a triangle abc, bc = 5, ca = 4, ab = 3 and d, e are the points on bc such that bd = de = ec

To find : Correct option

Solution:

ab = 3  ac = 4  bc = 5

5² = 4² + 3²

Hence Δ ABC is right angled triangle at A

bd = de = ec  =  bc/3 = 5/3

in Δ ACE

AE² = AC² + CE² - 2AC . CE Cos (∠C)

AC = 4  CE = 5/3  Cos (∠C) = 4/5  (  (AC/BC) as ABC is right angle triangle at A)

AE² = 4² + (5/3)² - 2(4)(5/3)(4/5)

=> AE² = 16 + 25/9  - 32/3

=> AE² = (144 + 25   - 96)/9

=> AE² =  73/9

=> AE = √73 / 3

CE² = AC² + AE² - 2AC . AE Cos∠CAE

=> (5/3)² = 4² + 73/9 - 2(4)(√73 / 3)Cos∠CAE

=> 25/9 = 16 + 73/9  - 8(√73 / 3)Cos∠CAE

=> 25 = 144 + 73 -  24√73Cos∠CAE

=> Cos∠CAE = 8/√73

=> Cos∠CAE = 8/√73

Sec∠CAE  = √73 /8

Sec²∠CAE  = 1 + tan²∠CAE

=> 73/64 = 1 + tan²∠CAE

=> 9/64 =  tan²∠CAE

=> tan∠CAE  = 3/8

option B & C are correct

AE² =  73/9

tan∠CAE  = 3/8

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