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tatiyna
3 years ago
8

Which additional information could be used to prove that the triangles are congruent using AAS or ASA? Check all that apply.

Mathematics
2 answers:
Ad libitum [116K]3 years ago
7 0

Consider the given triangles.

Given: \angle C \cong \angle Q

ASA congruence criterion states that if two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

AAS congruence criterion states that if two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Consider the first part:

1. In triangles ABC and QPT

If we take \angle B \cong \angle P , BC \cong PQ along with the given condition.

Then the triangles are congruent by ASA congruence criterion.

2. If we take \angle A \cong \angle T and AC= TQ=3.2 along with the given condition.

Then the triangles are congruent by ASA congruence criterion.

3. As,\angle C \cong \angle Q, \angle A \cong \angle T and \angle B \cong \angle P, so the triangles can not be congruent.

4. If we take \angle A \cong \angle T and BC \cong PQ along with the given condition.

Then the triangles are congruent by AAS congruence criterion.

5. If we take AC=TQ=3.2 and CB=QP=3.2 along with the given condition, then the triangles are congruent but by SAS congruence criteria neither by ASA nor AAS congruence criterion.

Ksju [112]3 years ago
7 0

Answer: it’s A,B, and D

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15) The diameter of a cylinder is 12 and height is 11. Find the surface area of the figure. ​
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3 years ago
A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

⇒ x = 4,19 miles

=> height = x tan 20° = 4,19 tan 20° = 1,525 miles

4) Calculate the hypotenuse of this triangle:

hipotenuese ² = x² + height ² = (4.19)² + (1.525)² = 19.88 miles²

hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.
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3 years ago
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