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In the triangle A B C, given the leg A B 325, the angle at A 33° 15', and the angle at C 56° 45'; to find the hypothenuse and the leg BC.

Making the hypothenușe radius, the proportions will be;

To find the hypothenuse. To find the leg BC. As sine of C, 560 45' 9.922355 As sine of C, 560 45' 9.922355 : leg A B, 325. 2.511883 : leg AB, 325 2.511883 :: Radius

10.000000 :: Sine of A, 33° 15' 9.739013

12.511883
9.922355

12.250896 9.922355

: Hyp. 388.6 2.5895281: leg. B, C. 2131 2.328541 NOTE. If the leg B C had been given, instead of the leg

A B, the proportions would have been the same, the obvious changes being made.

BY NATURAL SINES.

To solve this CASE by natural sines, institute the following proportions :

To find the hypothenuse. As THE NATURAL SINE OF THE ANGLE OPPOSITE THE GIVEN LEG, IS TO THE LENGTH OF THE LEG, SO IS UNITY, OR 1, TO THE LENGTH OF THE HYPOTHE. NUSE.

Or, which is the same thing, DIVIDE THE GIVEN LEG BY THE NATURAL SINE OF ITS OPPOSITE ANGLE, AND THE QUOTIENT WILL BE THE HYPOTHÉXUSE.

To find the other leg. As THE NATURAL SINE OF TIE AN. GLE OPPOSITE THE GIVEN LEG, IS TO THE LENGTH OF THE GIVEN LEG, SO IS THE NATURAL SINE OF THE ANGLE OPPOSITE THE OTHER LEG, TO THE LENGTH OF THE OTHER LEG.

EXAMPLE

Given leg 325. Nat. sine of 560 45', the angle opposite the given leg, 0.83629. Nat. sinė of 330 15', the angle opposite the other leg, 0.54829.

As 0.83629 : 325 ::1; 383.6.

CASE III.

Fig. 44.

The hypothenuse and one leg given, to find the angles and the other leg. Fig. 44.

A

B в

40 In the triangle A CB, given the hypothenuse A C 50, and the leg A B 40, to find the angles and leg B.C.

Making the hypothenuse radius, the proportion to find the angle A CB will be:

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The angle A C B being 53° 8', the other is consequently 36° 52'.

The angles being found, the leg B C. may be found by either of the preceding CASES.

It is 30.

BY NATURAL SINES.

The angle opposite the given leg, may be found by the following proportion:

As THE HYPOTHENUSE IS TO UNITY, OR 1, so IS THE GIVEN LEG TO THE NAT. SINE OF ITS OPPOSITE ANGLE.

Or, which is the same thing, DIVIDE THE GIVEN LEG BY THE HYPOTHENUSE, AND THE QUOTIENT WILL BE THE NAT. SINE.

EXAMPLE.

quotient 0.80000; which, looking in the table of nat. sines, the nearest corresponding number of degrees and minutes will be found to be 53° 8', the angle A C B.

BY THE SQUARE ROOT.

In this case the required leg may be found by the square root, without finding the angles; according to the following PROPOSITION:

IN EVERY RIGHT ANGLED TRIANGLE, THE SQUARE OF THE HYPOTHENUSE IS EQUAL TO THE SUM OF THE SQUARES OF THE TWO LEGS. Hence,

THE SQUARE OF THE GIVEN LEG BEING SUBSTRACTED FROM THE SQUARE OF THE HYPOTHENUSE, THE REMAINDER WILL BE THE SQUARE OF THE REQUIRED LEG.

As in the preceding EXAMPLE; the square of the leg A B 40 is 1600; this substracted frðin the square of the hypothenuse 50, which is 2500, leaves 900, the square of the leg BC, the square root of which is 30, the length of leg B C, as found by logarithms.

CASE IV.

Fig. 45.

с

89

The legs given, to find the angles and hypothenuse. Fig. 45.

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In the triangle A B C, given the leg A B 78.7, and the leg B C 89; to find the angles and hypothenuse.

Making the leg A B radius, the proportion to find the angle

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'The angle A CB is consequently 41° 29'. Making the leg B C radius, the proportion to find the angle B C A will be similar, with the obvious differences.

The angles being found, the hypothenuse may be found by CASE II. It is nearest 119.

BY THE SQUARE ROOT

In this case, the hypothenuse may be found by the square root, without finding the angles; according to the following PROPOSITION.

IN EVERY RIGHT ANGLED TRIANGLE, THE SUM OF THE SQUARES OF THE TWO LEGS IS EQUAL TO THE SQUARE OF THE HYPOTHENUSE.

In the above EXAMPLE, the square of A B 78.7 is 6193.69, the

square of B C 89 is 7921; these added make 14114.69, the

square root of which is nearest 119.

BY NATURAL SINES.

The hypothenuse being found by the square root, the angles

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