Question

AM = MB, ∠AOB = 140°

∠OAM =

पर्याय

• 50°

• 55°

• 60°

• 45°

Answer 1

55°

Explanation:

1. Known facts:

• AM = MB → M is the midpoint of chord AB. 
• ΔOAB is isosceles (OA = OB, both radii).
• ∠AOB = 140°.
• We want ∠OAM.

2. First, find ∠OAB in ΔOAB:

∠OAB = ∠OBA

= `(180^circ - 140^circ)/2`

= 20°

3. Now, locate M and think about ΔOAM: 

M lies on chord AB such that AM = MB.

This does not mean OM bisects ∠AOB, because M is not necessarily the midpoint of the arc only of the chord.

4. Use triangle properties:

In ΔOAM, side OA = radius r and AM is half of AB.

From ΔOAB, we can use the sine rule or cosine rule to get OM, then find ∠OAM.

5. Calculation:

In ΔOAB:

By cosine rule:

`AB = sqrt(OA^2 + OB^2 - 2 * OA * OB * cos 140^circ)`

= `sqrt(r^2 + r^2 - 2r^2 cos 140^circ)`

= `rsqrt(2 - 2 cos 140^circ)`

cos 140° = – cos 40° ≈ – 0.7660

So:

`AB = rsqrt(2 - 2(-0.7660))`

= `rsqrt(2 + 1.532)`

= `rsqrt(3.532) ≈ 1.879r`

Half of this:

`AM = (AB)/2 ≈ 0.9395r`

6. In ΔOAM:

Sides: OA = r, AM ≈ 0.9395r

We also know OM is from midpoint, so use cosine rule again:

OM² = OA² – AM² + (something?) (instead, easier to use sine rule in OAB)

Better: In ΔOAB, median OM length is given by Apollonius theorem:

`OM^2 = (OA^2 + OB^2)/2 - (AB)^2/4`

= `(r^2 + r^2)/2 - (1.879r)^2/4`

= `r^2 - (3.532r^2)/4`

= r² – 0.8832r²

= 0.117r²

OM ≈ 0.3417r

7. Now find ∠OAM in ΔOAM:

By sine rule:

`(sin ∠OAM)/(OM) = (sin ∠AOM)/(AM)`

We know ∠AOM = 70° (half of 140°), so:

`(sin ∠OAM)/(0.3417r) = (sin 70^circ)/(0.9395r)`

Cancel r:

`sin ∠OAM = 0.3417 * (sin 70^circ)/0.9395`

sin 70° ≈ 0.9397, so:

sin ∠OAM ≈ 0.3417 · 1.0002 ≈ 0.3418

This gives ∠OAM ≈ 55°.