EXAMPLES.

Ex. 1. In the triangle ABC are given AB = 123.5, the angle B = 39° 47′ 20′′, and C = 74° 52′ 10′′: required the

rest.

Construction.

The angle A = 180 − (B + C) = 180° — 114° 39′ 30′′ 65° 20' 30".

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Draw AB (Fig. 45) = 123.5. At the points A and B draw AC, BC, making the angles BAC and ABC equal, respectively, to 65° 20′ 30′′ and 39° 47′ 20′′; then will ABC be the triangle required.

Ex. 2. Given the side AB = 327, the side BC = 238, and the angle A = 32° 27', to determine the rest.

NOTE. It will be seen that in the above example the result is uncertain. The sine of an angle being equal to the sine of its supplement, it is impossible, from the sine alone, to determine whether the angle should be taken acute or obtuse. By reference to the construction, (Fig. 46,) we see that whenever the side opposite the given angle is less than the other given side, and greater than the perpendicular BD, the triangle will admit of two forms: ABC, in which the angle opposite to the side AB is acute, and ABC', in which it is obtuse. If BC were greater than BA, the point C' would fall on the other side of A, and be excluded by the conditions. If it were less than BD, the circle would not meet AC, and the question would be impossible.

Ex. 3. Given the side AB 37.25 chains, the side AC = 42.59 chains, and the angle C 57° 29′ 15′′, to determine the rest.

Ans. BC 32.774 chains, A = 47° 53′ 52′′, and B = 74° 36' 53".

Ex. 4. Given the angle A 29° 47' 29", the angle B = 24° 15' 17", and the side AB 325 yards, to find the other sides. Ans. AC = 164.93, BC 199.48.

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Ex. 5. The side AB of an obtuse-angled triangle is 127.54 yards, the side AC 106.49 yards, and the angle B 52° 27' 18", to determine the remaining angles and the side BC.

Ans. C 108° 16' 3", A 19° 16′ 39", BC = 44.34.

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Ex. 6. Given AB 527.63 yards, AC

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= 398.47 yards,

and the angle B 43° 29′ 11′′, to determine the rest.

Ans. C 65° 40′ 44′′, A = 70° 50′ 5′′, BC = 546.93;

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or, C = 114° 19′ 16′′, A = 22° 11′ 33′′, BC = 218.71.

CASE 2.

140. Two sides and the included angle being given, to determine the rest.

RULE 1.

Subtract the given angle from 180°: the remainder will be the sum of the remaining angles. Then,

As the sum of the given sides is to their difference, so is the tangent of half the sum of the remaining angles to the tangent of half their difference.

This half difference added to the half sum will give the angle opposite the greater side, and subtracted from the half sum will give the angle opposite the less side.

Then having the angles, the remaining side may be found by Case 1.

DEMONSTRATION.-The second paragraph of this rule may be enunciated in general terms; thus,

As the sum of two sides of a plane triangle is to their difference, so is the tangent of half the sum of the angles opposite those sides to the tangent of half the difference of those angles.

Let ABC (Fig. 47) be the triangle of which the side AC is greater than AB. With the centre A and radius AC describe a circle cutting AB produced in E and F. Join EC and CF, and draw Then, because ABC and AFC

FG parallel to BC.

Fig. 47.

G

E

A

B

since the half sum of two quantities taken from the greater leaves their half difference, CFG = EFG EFC = ABC — EFC = (ABC — ACB).

Now, since the angle ECF is an angle in a semicircle, it is a right angle. Therefore, if with the centre F and radius FC an arc be described, EC and CG will be the tangents of EFC and CFG, or of the half sum and half difference of ABC and ACB. But (2.6) EB: BF:: EC: CG.

Whence AC+AB: AC — AB :: tan. ≥ (ABC + ACB) : tan. } (ABC — ACB).

EXAMPLES.

Ex. 1. Given AB = 527 yards, AC = 493 yards, and the angle A 37° 49'.

Here

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C+B=180° - 37° 49′ = 142° 11', and

Ex. 2. In the triangle ABC are given AB=1025.57 yards, BC849.53 yards, and the angle B = 65° 43′ 20′′, to find the rest.

Ans. A 48° 52′ 10′′, C= 65° 24' 30", AC = 1028.13.

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Ex. 3. Two sides of a triangle are 155.96 feet and 217.43 feet, and their included angle 49° 19', to find the

rest.

Ans. Angles, 85° 4′ 12′′, 45° 36′ 48′′, side, 165.49.

RULE 2.

141. As the less of the two given sides is to the greater, so is radius to the tangent of an angle; and as radius is to the tangent of the excess of this angle above 45°, so is the tangent of the half sum of the opposite angles to the tangent of their half difference.

Having found the half difference, proceed as in Rule 1.

NOTE. This rule is rather shorter than the last, where the two sides have been found in a preceding calculation, and thus their logarithms are known.

DEMONSTRATION.-Let ABC (Fig. 48) be any plane triangle. Draw BD perpendicular to AB, the greater, and equal to BC, the less side.

BED = = 45°.

Make BE= BD, and join ED. Then, since BE = BD, the angle BDE; and since EBD is a right angle, BDE But BED+BDE = 2 BDE = BAD + BDA, and BDE = (BDA + BAD). But the half sum of any two quantities being taken from the greater will leave the half difference: therefore

ADE is the half difference of BDA and BAD.

Now, (Rule 3, Art. 137,) BD or BC: BA :: rad. : tan. ADB;

and (demonstration to last rule) AB+BD: AB-BD :: tan. † (BDA + BAD): tan. † (BDA— BAD):: tan. BDE: tan. ADE; but BDE being equal to 45°, its tangent =rad.

And ADE= (ADB — 45°) ... AB + BD: AB. -BD::r: tan. (ADB — 45o); but AB+ BC AB- BC: tan. (ACB+ BAC): tan. (ACB-BAC); whence r: tan. (ADB — 45°) :: tan. † (ACB+ BAC) : tan. † (ACB — BAC).

EXAMPLES.

Ex. 1. In the course of a calculation I have found the logarithm of AB = 2.596387, that of BC= 2.846392: now, the angle B being 55° 49', required the side AC.