Question

please answer with proper steps and valid process ​

Answer 1

$\underline{\large{\sf Answer :}}$

IN the given figure,

OBC and OKH are straight lines, AH = AK, b = 50° , c = 110° ---(Given)

Here, in triangle ACB,

/_ACB + /_CBA + /_BAC = 180°

--(sum of angles of triangle)

/_ACB = b = 50° + /_CBA = c = 110° + /_BAC = 180°

/_BAC = 180° - 160°

/_BAC = 20°

In triangle AHK, AH=AK therefore it is an isosceles triangle,

∴ /_AHK = /_AKH = x

/_AHK + /_AKH + /_KAH = 180°

x + x + /_KAH = 180°

/_KAH = /_BAC = 20°

∴ x + x + 20° = 180°

∴ 2x = 180° - 20°

x = (160°)/2

x = 80°= /_AHK = /_AKH

Now,

/_ABO = /_BAC + /_ACB

--------(Remote interior angles)

= 20° + 50°

/_ABO = 70°

/_AKH congruent to /_BKO = 80°

---(vertically opposite angles)

So, in triangle KOB,

/_KBO + /_BOK +/_BKO = 180°

/_KBO = /_ABO

∴ 70° + /_BOK + 80° = 180°

∴ /_BOK + 150° = 180°

/_BOK = 180° - 150°

/_BOK = d = 30°

Hence the value of d is option (c) 30°

Answer 2

Answer:

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