Question

Challenge for all MODs and Stars #2
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Answer 1

Solution :

Consider ∆ ABC .

AB = AC [ As ∆ABC is isosceles ] .

Angle BAC = 60° + 20° = 80°

Let Angle ABC = Angle ACB = x .

We know that sum of all angles of a triangle = 180°

=> x + x + 80° = 180°

=> 2x = 100°

=> x = 50°

Hence Angle ABC = Angle ACB = 50°

Angle PBC = 50° - 10° = 40°

In ∆ APB

Angle BAP + Angle ABP + Angle BPA = 180°

=> Angle BPA = 180° - [ 30° ]

=> Angle BPA = 150°

So , reflex angle BPA = 360° - 150°

=> 210°

We can say that x + Angle BPC = 210°

Now see ∆ BPC

It is known for sure that angle PBC = 40°

Suppose Angle ACP = k .

Then angle PCB becomes ( 50 - k ) .

Now we need to use the angle sum property in ∆ APC and ∆ BPC to get 2 equations with 2 variables x and k .

See ∆ APC

60 + x + k = 180°

=> x + k = 120°

x = 90 + k

=> 90 + 2k = 120°

=> 2k = 30°

=> k = 15°

x = 105°

Answer : The required value of x is 105° .

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