Question

in the ∆ABC ,angle abc= 60° , angle acb=45° and bc=10cm find the altitude AD of the triangle.

PLEASE DON'T GIVE SILLY ANSWERS IT'S URGENT! ​

Answer 1

Answer:

AD = 6.36 cm

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Step-by-step explanation:

Tan60° = AD/BD

=> √3 = AD/BD

=> BD = AD/√3 ……….(1)

Tan45° = AD/CD

=> 1 = AD/CD

=> 1 = AD/(10-BD)

=> 10-BD = AD

=> AD = 10- BD ………(2)

=> By (1) & (2)

AD/ √3 = 10-BD

=> AD = 10√3 - √3BD

=> AD +√3BD = 10√3

= AD(1+√3) = 10√3

=> AD = 10√3 / (1+√3)

= AD = {10√3( 1-√3)} /{(1+√3)(1-√3)}

= AD = (10√3 - 30) / -2

=> AD = (30–10√3) / 2

=> AD = 15 - 5√3

=> AD = 15 - 5*1.732

=> AD = 15- 8.66

= AD = 6.36 cm ( approx)

Answer 2

Given : In the ∆ABC , ∠ABC= 60° , ∠ACB =45° and BC=10cm

To find :  the altitude AD of the triangle.

Solution:

AD  is altitude Hence ΔABD and ΔACD are right angle triangles

D lies on BC

∠ABC= 60° => ∠ABD = 60°      and ∠ACD = ∠ACB = 45°

in ΔACD

Tan 45°  = AD / CD

=> 1 = AD/CD

=> CD = AD

in ΔABD

Tan 60°  = AD / BD

=> √3 = AD/BD

=> BD = AD/√3

CD + BD  = 10 cm

=> AD  +  AD/√3 = 10

=> AD  √3 +  AD = 10√3

=> AD = 10√3 / ( √3 + 1)

=> AD  = 10 √ 3 (√3 - 1) / (3 - 1)

=> AD  = 10   (3 - √3 ) / (2)

=> AD = 5 (3 - √3 )

=> AD ≈ 6.34 cm

the altitude AD of the triangle is  5 (3 - √3 )  cm   ≈ 6.34 cm

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