(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far f...
Question 1 Report
(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \(\pi = \frac{22}{7}\) and radius of earth = 6400km].
(b)
I In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.
Answer Details
(a) \(X\) is on longitude \(67^\circ E\) and \(Y\) on \(123^\circ E\); both lie on the equator. Take \(\pi=\frac{22}{7}\) and \(R=6400\text{ km}\).
(i) Distance between \(X\) and \(Y\) along the equator. The difference in longitude is
\[\theta=123^\circ-67^\circ=56^\circ.\]Distance along the equator (a great circle):
\[d=\frac{\theta}{360^\circ}\times 2\pi R=\frac{56}{360}\times 2\times\frac{22}{7}\times 6400.\]\[d=\frac{56}{360}\times\frac{2\times22\times6400}{7}=\frac{56}{360}\times 40228.57\]\[d\approx 6257.78\text{ km}\approx\mathbf{6258\text{ km}}.\](ii) Distance of \(X\) from the North pole. The North pole is \(90^\circ\) of latitude from the equator, measured along a meridian (a great circle):
\[d=\frac{90}{360}\times 2\pi R=\frac{1}{4}\times 2\times\frac{22}{7}\times 6400\]\[d=\frac{1}{4}\times 40228.57\approx\mathbf{10057\text{ km}}.\](b) In circle centre \(O\), \(N\) is the mid-point of chord \(PQ\) with \(|PQ|=8\text{ cm}\), \(|ON|=3\text{ cm}\) and \(\angle ONR=20^\circ\). Find \(\angle ORN\).
Since \(N\) is the mid-point of the chord, \(ON\perp PQ\) and
\[|PN|=\tfrac{1}{2}|PQ|=4\text{ cm}.\]The radius is \(|OP|\); by Pythagoras in right triangle \(ONP\):
\[|OP|=\sqrt{|ON|^2+|PN|^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.\]\(R\) lies on the circle, so \(|OR|=5\text{ cm}\) too. Now apply the sine rule in triangle \(ONR\), where \(|ON|=3\), \(|OR|=5\) and \(\angle ONR=20^\circ\):
\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]Download The App On Google Playstore
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