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Example. The top of a ship's mast is 663 ft. above the sea-level, and from it the lamp of a lighthouse can just be seen. After the ship has sailed directly towards the lighthouse for half-an-hour the lamp can be seen from the deck, which is 24 ft. above the sea. the rate at which the ship is sailing.

Let L denote the lamp, D and E the two positions of the ship, B the top of the mast, C the point on the deck from which the lamp is seen; then LCB is a tangent to the earth's surface at A.

[In problems like this some of the lines must necessarily be greatly out of proportion.] Let AB and AC be expressed in miles; then since DB=66 feet and EC 24 feet, we have by the rule

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2×663=100;

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But the angles subtended by AB and AC at O the centre of the earth are very small;

.. arc AD=AB, and arc AC=AE.

.. arc DEAD-AE=AB-AC-4 miles.

[Art. 268.]

Thus the ship sails 4 miles in half-an-hour, or 8 miles per hour.

275. Let be the number of radians in the dip of the horizon; then with the figure of Art. 273, we have

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neglect the terms on the right after the first.

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Let N be the number of degrees in 6 radians; then

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Now r=63 nearly; hence we have approximately

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a formula connecting the dip of the horizon in degrees and the height of the place of observation in miles.

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1. Find the greatest distance at which the lamp of a lighthouse can be seen, the light being 96 feet above the sealevel.

2. If the lamp of a lighthouse begins to be seen at a distance of 15 miles, find its height above the sea-level.

3. The tops of the masts of two ships are 32 ft. 8 in. and 42 ft. 8 in. above the sea-level: find the greatest distance at which one mast can be seen from the other.

4. Find the height of a ship's mast which is just visible at a distance of 20 miles from a point on the mast of another ship which is 54 ft. above the sea-level.

5. From the mast of a ship 73 ft. 6 in. high the lamp of a lighthouse is just visible at a distance of 28 miles: find the height of the lamp.

6. Find the sexagesimal measure of the dip of the horizon from a hill 2640 feet high.

7. Along a straight coast there are lighthouses at intervals of 24 miles: find at what height the lamp must be placed so that the light of one at least may be visible at a distance of 3 miles from any point of the coast.

8. From the top of a mountain the dip of the horizon is 1·81°: find its height in feet.、

9. The distance of the horizon as seen from the top of a hill is 30.25 miles: find the height of the hill and the dip of the horizon.

10. If x miles be the distance of the visible horizon and N degrees the dip, shew that

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14. Two sides of a triangle are 31 and 32, and they include a right angle: find the other angles.

15. A person walks directly towards a distant object P, and observes that at the three points A, B, C, the elevations of P are a, 2a, 3a respectively: shew that AB=3BC nearly.

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CHAPTER XXII.

GEOMETRICAL PROOFS.

276. To find the expansion of tan (A+B) geometrically. Let ▲ LOM=A, and ▲ MON=B; then ▲ LON=A+B.

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In ON take any point P, and draw PQ and PR perpendicular to OL and OM respectively. Also draw RS and RT perpendicular to OL and PQ respectively.

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also the triangles ROS and TPR are similar, and therefore

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In like manner, with the help of the figure on page 95, we may obtain the expansion of tan (A-B) geometrically.

277. To prove geometrically the formula for transformation of sums into products.

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Let EOF be denoted by A, and LEOG by B.

With centre O and any radius describe an arc of a circle meeting OG in H and OF in K.

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Bisect

KOH by OL; then OL bisects HK at right angles.

Draw KP, HQ, LR perpendicular to OE, and through L draw MLN parallel to OE meeting KP in M and QH in N.

It is easy to prove that the triangles MKL and NHL are equal in all respects, so that KM=NH, ML=LN, PR=RQ.

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