In the diagram, a ladder TF, 10 metres long is... | WAEC SSCE - General Mathematics - 1999

Question 1 Report

In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.

(a) How high up the wall, correct to the nearest metre, does the ladder reach?

(b) If the foot (F) of the ladder is pulled from the wall to F\(^{1}\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\(^{1}\).

(ii) Calculate, correct to the nearest degree, \(QF^{1}T^{1}\).

Answer

Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.

(a) Height reached up the wall. In right triangle \(TQF\):

\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\]

(b)(i) Distance the top slides down. First the original foot distance:

\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]

New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):

\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]

Distance slid down:

\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\]

(b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):

\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]

Read lesson note on Introductory Calculus (WAEC)

Read lesson note on Circles (WAEC)

Read lesson note on Statistics (WAEC)

Answer Details

(a) Height reached up the wall. In right triangle \(TQF\):

\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\]

(b)(i) Distance the top slides down. First the original foot distance:

\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]

New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):

\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]

Distance slid down:

\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\]

(b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):

\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]

Read lesson note on Introductory Calculus (WAEC)

Read lesson note on Circles (WAEC)

Read lesson note on Statistics (WAEC)

Introductory Calculus

Circles

Statistics

View Answer

In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal. (a) How high up the wall, correct to the nearest ...

Download The App On Google Playstore

In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal. (a) How high up the wall, correct to the nearest ...

Related Questions

Download The App On Google Playstore