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aivan3 [116]
3 years ago
12

A student writes an incorrect step while checking if the sum of the measures of the two remote interior angles of triangle ABC b

elow is equal to the measure of the exterior angle. Step 1: m∠m + m∠n + m∠o = 180 degrees (sum of angles of a triangle) Step 2: m∠p − m∠o = 90 degrees (alternate interior angles) Step 3: Therefore, m∠m + m∠n + m∠o = m∠o + m∠p Step 4: So, m∠m + m∠n = m∠p In which step did the student first make a mistake and how can it be corrected? Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (corresponding angles) Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (adjacent angles) Step 2; it should be m∠o + m∠p = 180 degrees (alternate exterior angles) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)
I know its one of the step two answers but which one?
Mathematics
1 answer:
Goshia [24]3 years ago
4 0

Given steps  :

Step 1: m∠m + m∠n + m∠o = 180 degrees (sum of angles of a triangle)

Step 2: m∠p − m∠o = 90 degrees (alternate interior angles) : <em>It's incorrect step because m<p and m<0 are on a common point on a line and make a linear pair. Therefore, m<p and m<0 are supplementary angles.</em>

Step 3: Therefore, m∠m + m∠n + m∠o = m∠o + m∠p.

Step 4: So, m∠m + m∠n = m∠p.

<h3>Therefore, student did mistake in 2nd step and correct step should be  Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles).</h3>
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As per the given information;

m<E = 22, the triangle EDF (the one that is given in the picture) is isosceles (meaning that the sides are congruent).

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To keep it simple while solving the problem, let's say that they have a value of x degrees.

So,

m<E + m<D + m<F =180

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The length of arc EF = $2 \pi r \times \frac{\theta}{360^\circ}$

                                   $=2 \times 3.14 \times 6 \times \frac{140^\circ}{360^\circ}$

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In circle C

\angle GCH = \theta = 140^\circ

The length of arc GH = $2 \pi r \times \frac{\theta}{360^\circ}$

                                   $=2 \times 3.14 \times 4 \times \frac{140^\circ}{360^\circ}$

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The length of EF is  1.5 times the length of GH

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