A radioactive isotope that was originally 275 grams has decayed to 150 grams. the equation shown can be used to calculate the nu
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The values of the angles are; x = 12°, m∠PQS = 71°, m∠PQT = 142° and m∠TQR = 41°
<h3>What are the measure of the each angle?</h3>The required angles; x, m∠PQS, m∠PQT, and m∠TQR
The given parameters are bisects ∠PQT
m∠SQT = (8·x - 25)°
m∠PQT = (9·x + 34)°
m∠SQR = 112°
We have;
m∠PQT = m∠SQT + m∠PQS (Angle addition postulate)
m∠SQT ≅ m∠PQS (Angles formed by angle bisector are congruent)
m∠SQT = m∠PQS by Definition of congruency
m∠PQT = 2 × m∠SQT
Therefore;
(9·x + 34)° = 2 × (8·x - 25)° = (16·x - 50)°
Collecting like terms gives:
(34 + 50)° = 16·x - 9·x = 7·x
7·x = 84°
x = 84°/7 = 12°
x = 12°
m∠SQT = (8·x - 25)°
Therefore;
m∠SQT = (8 × 12 - 25)° = 71°
m∠SQT = 71°
m∠PQT = 2 × m∠SQT
∴ m∠PQT = 2 × 71° = 142°
m∠PQT = 142°
m∠PQS = m∠SQT (Angles formed by the same bisector )
∴ m∠PQS = m∠SQT = 71°
m∠PQS = 71°
m∠SQR = m∠SQT + m∠TQR (Angle addition postulate)
m∠SQT = 71°
∴ m∠SQR = 112° = 71° + m∠TQR
m∠TQR = 112° - 71° = 41°
m∠TQR = 41°
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ANSWER
EXPLANATION
To find a positive angle that is coterminal with
We add multiples of 2π until we get a positive angle that is less than one revolution,
We add to obtain,
This simplifies to,
This is the first positive angle that is coterminal with
and is less than one revolution.
EXPLANATION
To find a positive angle that is coterminal with
We add multiples of 2π until we get a positive angle that is less than one revolution,
We add to obtain,
This simplifies to,
This is the first positive angle that is coterminal with
and is less than one revolution.