Vipin S.

asked • 12/31/17

In a right angled triangle ABC. Right angled at A, angle B is 40°.if M is the mid point of hypotenuse BC.then angle AMC=?

Dear sir please explain the above problem.
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Muhammad H.

∠A=900 ,∠B=400,so ∠c=500.
Now M is the midpoint of the hypotenuse BC.We know that the  pericentre of a right angled triangle is the midpoint of the hypotenuse(it has a easy proof!).
So M is the pericentre of the circle ABC and so, CM=BM=AM (they are the radius of the ABC circle).
So, in the ΔAMC,
AM=CM
So,∠MAC=∠C=500.
And ∠AMC=1800-(∠MAC+∠C)
               =1800-(500+500)
               =800 (Answer)
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01/01/18

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Paul H. answered • 01/01/18

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The segment AM creates 2 triangles, ABM and ACM, each of which is isosceles. AB becomes the base for triangle ABM, and AC becomes the base for triangle ACM. Since the base angles of an isosceles triangle are congruent, angle ABM (40-degrees) is the same as angle BAM, and angle ACM (50-degrees) is the same as angle CAM. Angle AMC is the vertex angle of an isosceles triangle with base angles of 50-degrees each, so angle AMC is 180 - (50 x 2) = 80-degrees.
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Paul H.

If you're wondering how we know that triangles ABM and ACM are isosceles, think backwards. First, use AC as the base and draw a line from point A to point M, such that M is on line BC and angle CAM measures the same as angle ACM (which is 50-degrees). This creates a triangle with congruent base angles, which means that the sides opposite these base angles are congruent. So we have AM = CM. We also know that angle BAM measures 90-degrees minus 40-degrees, which equals 50-degrees, which is the same as angle ABM. Thus, AB is the base of triangles ABM, and the base angles of this triangle are congruent (each is 50-degrees). Since the sides opposite the base angles of an isosceles triangle are congruent, we know that BM = AM = CM, and hence M is the midpoint of BC.
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01/01/18

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