Angles A, B and C of a triangle are in arithmetic progression. M is a point on BC such that AM is perpendicular to BC.
What is $$\frac{BM}{AB}$$​?

Correct option is A

Given:

Angles A, B and C of a triangle are in arithmetic progression

​$$AM \perp BC$$​​

Formula Used:

Pythagoras’ Theorem

​$$\cos \theta = \frac{Base}{Hypotenuse}$$​​

​$$\cos 60^o = \frac{1}{2}$$​​

Solution:

​$$A + B +C = 180^o$$​     …(1)

If A, B and C are in AP then

B- A = C -B

2B = A +C    ….(2)

Solving eq(1) and we get B = $$60^o$$​

AM is perpendicular to BC hence $$\triangle ABM$$ is a right angled triangle and $$\angle AMB = 90^o$$​​

The remaining $$\angle BAM = 30^o$$​  

​$$\cos 60^o = \frac{BM}{AB} = \frac{1}{2}$$​​

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