To the nearest whole number, what is the surface area of a sphere with a volume of 140 cm³?
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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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For the division part:
But I think your question differs than this..
Am I right?
as your options do not include 1
But I think your question differs than this..
Am I right?
as your options do not include 1