(a)
In the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \(36 cm^{2}\). Find the value of |QR|.
(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.
In the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \(36 cm^{2}\). Find the value of |QR|.
(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.
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Correct Answer: Option n
Explanation:
(a)
In the diagram above, area of \(\Delta PSR = \frac{1}{2} \times |PR| \times 6\)
\(36 = \frac{1}{2} \times |PR| \times 6\)
\(36 = 3 \times |PR|\)
\(|PR| = 12 cm\)
But |PQ| = |TS| = 8cm (opp. sides of a parallelogram)
Hence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.
(b)
In the diagram above, \(\beta = 60°\) (alternate angles)
\(\tan \beta = \frac{10.65}{|QB|}\)
\(|QB| = \frac{10.65}{\tan 60}\)
= \(\frac{10.65}{1.732} = 6.149m\)
|PO| = |QB| = 6.149m (opp. sides of a rectangle)
In \(\Delta POT, \tan 45 = \frac{|TO|}{6.149}\)
\(|TO| = \tan 45 \times 6.149 = 1 \times 6.149 = 6.149m\)
|OB| = |PQ| = 10.65m (opp sides of a rectangle)
\(h = |OB| - |OT| \)
= \(10.65 - 6.149\)
= \(4.501 m \approxeq 4.50m\) (3 sig. figs)
Hence, the height of the flagpole = 4.50m
(a)
In the diagram above, area of \(\Delta PSR = \frac{1}{2} \times |PR| \times 6\)
\(36 = \frac{1}{2} \times |PR| \times 6\)
\(36 = 3 \times |PR|\)
\(|PR| = 12 cm\)
But |PQ| = |TS| = 8cm (opp. sides of a parallelogram)
Hence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.
(b)
In the diagram above, \(\beta = 60°\) (alternate angles)
\(\tan \beta = \frac{10.65}{|QB|}\)
\(|QB| = \frac{10.65}{\tan 60}\)
= \(\frac{10.65}{1.732} = 6.149m\)
|PO| = |QB| = 6.149m (opp. sides of a rectangle)
In \(\Delta POT, \tan 45 = \frac{|TO|}{6.149}\)
\(|TO| = \tan 45 \times 6.149 = 1 \times 6.149 = 6.149m\)
|OB| = |PQ| = 10.65m (opp sides of a rectangle)
\(h = |OB| - |OT| \)
= \(10.65 - 6.149\)
= \(4.501 m \approxeq 4.50m\) (3 sig. figs)
Hence, the height of the flagpole = 4.50m