Question

how can b construct a Abc m<A=45,M < B=60,M<C=75​

Answer 1

Answer:

IT IS NOT POSSIBLE

Step-by-step explanation:

Answer 2

AD is perpendicular to BC. & CF is perpendicular to AB. < DAB = 30° .

Since < A = 45° => < ACF = 45°

=> AF = CF = a

In triangle AFO

cos 30° = a/AO

=> √3/2 = a/ AO

=> AO = 2a/√3 …………..(1)

In triangle CFB

tan 60° = CF/FB

=> √3 = a/ FB

=> FB = a/√3 ………………(2)

So, AB = a + a/√3

=> AB = (√3 + 1)a / √3 ……………..(3)

Now, in trisngle ADB

Sin60° = AD/AB

=> √3/2 = AD / { (√3+1)a /√3} ( by eq 3)

=> 2 AD = (√3+1)a

=> AD = (√3+1)a /2 ………………(4)

=> OD = AD - AO = eq(4) - eq(1)

=> OD = (√3+1)a /2 - 2a/√3

=> OD = (3a + √3a - 4a)/2√3

=> OD = (√3a - a)/2√3 ………….(5)

Now, eq (1) ÷ eq(5)

=> AO/OD = (2a/√3) x ( 2√3/ √3a- a )

=> AO /OD = 4/(√3–1)

=> AO/OD = 4(√3+1) / 2

=> AO/OD = 2(√3+1) /2

So, orthocentre O divides altitude AD in the following ratio

=> AO/OD = 2(√3+1) : 1

OR = approximately 5.5 :1