(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.
(b)
In the diagram, PQR is a straight line, \(\overline{QR} = \sqrt{3} cm\) and \(\overline{SQ} = 2 cm\). Calculate, correct to one decimal place, < PQS.
(b)
In the diagram, PQR is a straight line, \(\overline{QR} = \sqrt{3} cm\) and \(\overline{SQ} = 2 cm\). Calculate, correct to one decimal place, < PQS.
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Correct Answer: Option
Explanation:
The sum of exterior angles = 360°
\(60° + 2(90 - m)° + 2(30 + 2m)° = 360°\)
\(60° + 180° - 2m° + 60° + 4m° = 360°\)
\(300° + 2m = 360°\)
\(\implies 2m = 360° - 300° = 60°\)
\(m = 30°\)
(b)
In \(\Delta QRS, \cos Q = \frac{\sqrt{3}}{2}\)
\(\cos Q = 0.8660 \implies Q = 30°\)
\(\therefore < SQR = 30°^\)
\(< PQS = 180° - 30° \) (angles on a straight line = 180°)
\(< PQS = 150°\)
The sum of exterior angles = 360°
\(60° + 2(90 - m)° + 2(30 + 2m)° = 360°\)
\(60° + 180° - 2m° + 60° + 4m° = 360°\)
\(300° + 2m = 360°\)
\(\implies 2m = 360° - 300° = 60°\)
\(m = 30°\)
(b)
In \(\Delta QRS, \cos Q = \frac{\sqrt{3}}{2}\)
\(\cos Q = 0.8660 \implies Q = 30°\)
\(\therefore < SQR = 30°^\)
\(< PQS = 180° - 30° \) (angles on a straight line = 180°)
\(< PQS = 150°\)