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

Making the hypothenuse radius, the proportions will be;

To find the hypothenuse. As sine BCA, 56° 45′

: leg AB, 325

: : radius

: hyp. 388.6

To find the leg BC.

As sine BCA, 56° 45'

|: leg AB, 325

|:: sine BAC, 33° 15'

leg BC, 213.1

NOTE. If the leg BC had been given, instead of the leg AB, the proportions would have been the same, the obvious changes being made.

BY NATURAL S INES.

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 HYPOTHENUSE.

To find the other leg. AS THE NATURAL SINE OF THE 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 56° 45', the angle opposite the given leg 0.83629. Nat. sine of 33° 15', the angle opposite the other leg 0.54829.

As 0.83629 : 325 : : 1 : 388.6.

CASE III.

Fig. 44.

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

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In the triangle ABC, given the hypothenuse AC 50 and the leg AB 40, to find the angles and leg BC.

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

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The angle ACB being 53° 8' the other is consequently 36° 52'.

The angles being found, the leg BC 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 looked in the table of nat. sines, the nearest corresponding number of degrees and minutes will be found to be 53° 8', the angle ACB.

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

THE TWO LEGS. HENCE,

OF

THE SQUARE OF THE GIVEN LEG BEING SUBTRACTED 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 AB 40 is 1600; this subtracted from 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 BC as found by logarithms.

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In the triangle ABC, given the leg AB 78.7 and the leg BC 89; to find the angles and hypothenuse.

Making the leg AB radius, the proportion to find the angle

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The angle ACB is consequently 41° 29'.

Making the leg BC radius, the proportion to find the angle BCA 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 AB 78.7 is 6193.69, the square of BC 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 may be found by nat. sines, according to the preceding

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