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When two sides of a right-angled triangle are given, the other side may be found by the following rules, without first finding the angles.
1. When the hypothenuse and one leg are given, to find the other leg.
Subtract the square of the given leg from the square of the hypothenuse; the square root of the remainder will be the leg required.* Or by logarithms thus,
To the logarithm of the sum of the hypothenuse and given side, add the logarithm of their difference; half this sum will be the logarithm of the leg required.
2. When the two legs are given to find the hypothenuse.
Add together the squares of the two given legs; the square root of the sum will be the hypothenuse.* Or by logarithms thus,
* DEMONSTRATION. The square of the hypothenuse of a right-angled tri angle is equal to the sum of the squares of the sides (47.1). Therefore the truth of the first part of each of the rules, is evident.
Put h = the hypothenuse, b = the base, and the perpendicular, then (47.1) p2 = h2 b2 (5.2. cor.) h + bx hb, or p=
h+bxh-b; whence from the nature of logarithms, the latter part of the first rule is evident.
which solved by logarithms will correspond with the latter part of
From twice the logarithm of the perpendicular, subtract the logarithm of the base, and add the corresponding natural number to the base; then, half the sum of the logarithms of this sum, and of the base, will be the logarithm of the hypothenuse.
1. The hypothenuse of a right angled triangle is 272, and the base 232; required the perpendicular.
2. Given the base 186, and the perpendicular 152, to
3. The hypothenuse being given equal 403, and one leg 321; required the other leg. Ans. 243.7.
4. What is the hypothenuse of a right-angled triangle, the base of which is 31.04, and perpendicular 27.2. Ans. 41.27.
The following examples, in which trigonometry is applied to the mensuration of inaccessible distances and heights, will serve to render the student expert in solving the different cases, and also to elucidate its use.
The Application of Plane Trigonometry to the Mensuration of Distances and Heights.
EXAMPLE 1. Fig. 54.
Being on one side of a river and wanting to know the distance to a house on the other side, I measured 500 yards along the side of the river in a right line AB, and found the two angles* between this line and the object to be CAB = 74° 14', and CBA Required the distance between each station and the object.
The sum of the angles CAB and CBA is 123° 37', which subtracted from 180° leaves the angle ACB = 56° 23. Then by case 1;
The angles may be taken with a surveyor's compass or any other similar
Suppose I want to know the distance between two places A and B, accessible at both ends of the line AB, and that I measured AC 735 yards, and BC= 840; also the angle ACB 55° 40'. What is the distance between A and B?
The angle ACB = 55° 40', being subtracted from 180°, leaves 124° 20′; the half of which is 62° 10′. Then by case 3.
= 7° 12′ and we shall have CAB = 69° 22′, and CBA≈ 54° 58′. Then,
Wanting to know the distance between two inaccessible objects A and B, I measured a base line CD = 300 yards: at C the angle BCD was 58° 20′ and ACD 95° 20'; at D the angle CDA was 53° 30′ and CDB 98° 45'.
1. In the triangle ACD, are given the angle ACD → 95° 20′, ADC = 53° 30′, and the side CD
300, to find
300, to find BC
2. In the triangle BCD, are given the angle BCD 58° 20′, BDC = 98° 45', and side CD = 761.47.
3. In the triangle ACB we have now given the angle ACB ACD - BCD = 37°, the side AC465.98 and BC = 761.47, to find AB
479.8 yards, the dis
EXAMPLE 4. Fig. 57.
Being on one side of a river and observing three objects A, B and C stand on the other side, whose distances apart 1 knew to be, AB = 3 miles, AC = 2, and BC= 1.8, I took a station D, in a straight line with the objects A and C, being nearer the former, and found the angle ADB = 17° 47'. Required my distance from each
of the objects.
With the three given distances, describe the triangle ABC; bisect BC in F and draw FE perpendicular to it; draw CE making the angle BCE 72° 13′ the complement of the given angle ADB; with E as a centre and distance EC, describe the circle BCD, meeting CA produced in D: then AD, CD and BD will be the distances required.*
DEMONSTRATION. By construction the distances AB, BC and AC are