(a) A regular polygon of n sides is such that each... | WAEC SSCE - General Mathematics - 2005 (Essay)

Question 1 Report

(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :

(i) the value of n ; (ii) the sum of all the interior angles.

(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.

(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.

Answer

(a) The regular polygon

(i) Value of n. At any vertex the interior and exterior angles are supplementary, and here the interior angle is 120° greater than the exterior angle. Let the exterior angle be \(x\):

\[x + (x + 120^\circ) = 180^\circ\] \[2x + 120^\circ = 180^\circ \implies 2x = 60^\circ \implies x = 30^\circ.\]

For any polygon the exterior angles sum to \(360^\circ\), so

\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{30^\circ} = 12.\]

The polygon has 12 sides.

(ii) Sum of all the interior angles.

\[\text{Sum} = (n-2)\times 180^\circ = (12-2)\times 180^\circ = 10 \times 180^\circ = 1800^\circ.\]

The interior angles sum to 1800°.

(b) The boy's journey

(i) Sketch of the movement. He walks 6 km from \(P\) to \(Q\) on a bearing of 065°, then 13 km from \(Q\) to \(R\) on a bearing of 146°. A North line is drawn at each point so the bearings can be measured clockwise from North.

Sketch of the boy's route: P to Q (6 km, bearing 065°) then Q to R (13 km, bearing 146°), with North lines at P and Q and the 99° interior angle at Q. The dashed line PR is the distance calculated.

(ii) Distance PR. First find the interior angle of the triangle at \(Q\). The bearing of \(P\) from \(Q\) is \(065^\circ + 180^\circ = 245^\circ\), and the bearing of \(R\) from \(Q\) is \(146^\circ\). Working from the North line at \(Q\), the angle between \(QP\) and the North line is \(180^\circ - 65^\circ = 115^\circ\) on the western side, giving the angle inside the triangle:

\[\angle PQR = (180^\circ - 146^\circ) + 65^\circ = 34^\circ + 65^\circ = 99^\circ.\]

Apply the cosine rule to triangle \(PQR\):

\[|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2\,|PQ|\,|QR|\cos(\angle PQR)\] \[|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13)\cos 99^\circ\] \[|PR|^{2} = 36 + 169 - 156\cos 99^\circ\]

Since \(\cos 99^\circ = -0.1564\):

\[|PR|^{2} = 205 - 156(-0.1564) = 205 + 24.40 = 229.40\] \[|PR| = \sqrt{229.40} = 15.15\text{ km}.\]

Correct to the nearest kilometre, \(|PR| \approx \mathbf{15\text{ km}}\).

Read lesson note on Triangles And Polygons (WAEC)

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