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As the side made radius,
After finding the angle, the other side is found as in the preceding rule.
I. In a right-angled triangle ABC, are given the base AB= 208, and the angle A = 35° 16', to find the hypothenuse AC and perpendicular BC.
The base AB being radius.
The perpendicular BC being radius.
2. In a right-angled triangle ABC, there are given the hypothenuse AC = 272, and the base AB= 232; re
2.43457 Is to AB 232
2.365-19 So is radius
The base AB being radius. As AB 232
2.36549 Is to AC 272
2.43457 So is radius
3. In a right-angled triangle, are given the hypothe.
36.57, and the Angle A = 27° 46', to find the base AB, and perpendicular BC. Ans. Base AB = 32.36, and perpendicular BC= 17.04.
4. In a right-angled triangle, there are given, the perpendicular = 193.6, and the angle opposite the base 47° 51'; required the hypothenuse and base. Ans. Hypothenuse = 288.5, and base = 213.9.
5. Required the angles and hypothenuse of a rightangled triangle, the base of which is 46.72, and perpendicular 57.9. Ans. Angle opposite the base 38° 54', angle opposite the perpendicular 51° 6', and hypothenuse 74.4.
When two sides of a right-angled triangle are given, the other side may be found by the following rules, with. out 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.
* DEMONSTRATION. The square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the sides (47.1).
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,
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.
* Put h= the hypothenuse, b the base, and p = the perpendicular, then (47.1) p2 =h2 - 62=(5.2. cor.) h + bxh - b, or p
12 . , = h + 6 x h 6; whence from the nature of logarithms, the latter part of the first rule is evident.
12 Also (47.1.) h2 = 12 + 112 = 6X6 + hv|
6 which solved by logarithms will correspond with the latter part of the second rule.
) 2 =
or h=v(o xo + )