Hello!
Since the angles are alternate interior angles they equal each other
5x - 28 = 3x + 12
Now you solve this algebraically
5x - 28 = 3x + 12
Subtract 3x from both sides
2x - 28 = 12
Add 28 to both sides
2x = 40
Divide both sides by 2
x = 20
Now we put x into the equation we are wanting to find out
5x - 28
5(20) - 28
Multiply
100 - 28
Subtract
72
The answers are x = 20 and 5x - 28 = 72°
Hope this helps!
Answer:
Step-by-step explanation:
The vector in component form is the difference of the coordinates of its end points:
rock - ship = (4, 7) - (1, 0) = (3, 7) . . . . vector from ship to rock
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The angle from North can be found using the tangent relation:
tan(bearing) = (east distance)/(north distance) = 3/7
bearing = arctan(3/7) ≈ 23.1986°
The bearing of the rock is 23.20°.
Answer:
Δ QRS ≈ Δ QST ≈ Δ SRT ⇒ 3rd answer
Step-by-step explanation:
From the given figure
In Δ QRS
∵ m∠S = 90°
∵ m∠S = m∠QST + m∠RST
∴ m∠QST + m∠RST = 90° ⇒ (1)
- Use the fact the sum of the measures of the interior angles
of a Δ is 180°
∴ m∠Q + m∠S + m∠R = 180°
∵ m∠S = 90
∴ m∠Q + 90° + m∠R = 180°
- Subtract 90 from both sides
∴ m∠Q + m∠R = 90° ⇒ (2)
In Δ QST
∵ m∠QTS = 90°
- By using the fact above
∴ m∠Q + m∠QST = 90 ⇒ (3)
- From (1) and (3)
∴ m∠QST + m∠RST = m∠Q + m∠QST
- Subtract m∠QST from both sides
∴ m∠RST = m∠Q
In Δ SRT
∵ m∠STR = 90°
- By using the fact above
∴ m∠R + m∠RST = 90 ⇒ (4)
- From (1) and (4)
∴ m∠QST + m∠RST = m∠R + m∠RST
- Subtract m∠RST from both sides
∴ m∠QST = m∠R
In Δs QRS and QST
∵ m∠S = m∠QTS ⇒ right angles
∵ m∠R = m∠QST ⇒ proved
∵ ∠Q is a common angle in the two Δs
∴ Δ QRS ≈ Δ QST ⇒ AAA postulate of similarity
In Δs QRS and SRT
∵ m∠S = m∠STR ⇒ right angles
∵ m∠Q = m∠RST ⇒ proved
∵ ∠R is a common angle in the two Δs
∴ Δ QRS ≈ Δ SRT ⇒ AAA postulate of similarity
If two triangles are similar to one triangle, then the 3 triangles are similar
∵ Δ QRS ≈ Δ QST
∵ Δ QRS ≈ Δ SRT
∴ Δ QRS ≈ Δ QST ≈ Δ SRT
Answer:
1,2,3
Step-by-step explanation:
Answer:
No, he should have set the sum of ∠AED and ∠DEC equal to 180°, rather then setting ∠AED and ∠DEC equal to each other.
Step-by-step explanation:
He was using the measurements of m∠AED & m∠CED, which are supplementary angles, not vertical angles (therefore making them, when combined, equal to 180°).
If they were vertical angles (at the case of m∠AED & m∠BEC, or the other pair), then yes, they will be congruent. But in this case, they are not, so you don't solve it like they are vertical angles.
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