in the triangle abc angle abc equal to 60 degree and angle ACB = 45 degree and BC = 10 cm find the altitude ad of the triangle
Answers
Step-by-step explanation:
Let the perpendicular be CD on AB.
In triangle ADC
tan60=CD/AD = 3^1/2
CD = AD*3^1/2 ………………..(1)
tan45 = CD/BD = 1
CD = BD ………………(2)
From (1) and (2)
we get AD*3^1/2 = BD
and AD + BD = 10cm
so AD(3^1/2 + 1) = 10cm
AD = 5(3^1/2 -1)
AD = 5*0.732 = 3.66cm
CD = AD*3^1/2
CD = 3.66*1.732 = 6.33912cm
hope it's help u ‼️✌️✌️
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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