The difference of two sides of a triangle is to their sum as the tangent of half the difference of the opposite angles is to the tangent of half their cum.

NOTE. If in [27] ba, then B>A.

The formula is still true, but to

avoid negative quantities, the formula in this case should be written

1. What do the formulas of § 33 become when one of the angles is a right angle?

2. Prove by means of the Law of Sines that the bisector of an angle of a triangle divides the opposite side into parts proportional to the adjacent sides.

3. What does Formula [26] become when A=90° ? when A=0°? when A=180°? What does the triangle become in each of these cases?

NOTE. The case where A = 90° explains why the theorem of § 34 is sometimes termed the Generalized Theorem of Pythagoras.

4. Prove (Figs. 31 and 32) that whether the angle B is acute or obtuse, c = a cos B +b cos A. What are the two symmetrical formulas obtained by changing the letters? does the formula become when B=90° ?

What

5. From the three following equations (found in the last exercise) prove the theorem of § 34:

c = a cos B+b cos A,

b =a cos C+c cos A,
a=b cos C+c cos B.

HINT. Multiply the first equation by c, the second by b, the third by a; then from the first subtract the sum of the second and third.

2

6. In Formula [27] what is the maximum value of (A-B)? 7. Find the form to which Formula [27] reduces, and describe the nature of the triangle, when

(i.) C'=90°;

(ii) A-B=90°, and B= C.

§ 36. THE SOLUTION OF AN OBLIQUE TRIANGLE. The formulas established in §§ 33-35, together with the equation A+B+C=180°, are sufficient for solving every case of an oblique triangle. The three parts that determine an oblique triangle may be:

I. One side and two angles;

II. Two sides and the angle opposite to one of these sides; III. Two sides and the included angle;

IV. The three sides.

Let A, B, C denote the angles, a, b, c the sides respectively.

§ 37. CASE I.

Given one side a, and two angles A and B; find the remaining parts C, b, and c.

EXAMPLE. α=24.31, A=45° 18', B=22° 11'.

The work may be arranged as follows:

NOTE. When 10 is omitted after a logarithm or cologarithm, it must

be remembered that the log or colog is 10 too large.

EXERCISE XVII.

1. Given a= = 500,
find C=123° 12',

2. Given a=795,

find C55° 20',

3. Given a = = 804,

find C-35° 4', 4. Given a=820,

find C=25° 12', 5. Given c=1005,

find C = 47° 14′,

6. Given b=13.57,

find A=108° 50',

7. Given a=6412,

find B=56° 56',

8. Given b=999,

find B=77°,

9. In order to determine the distance of a hostile fort A from a place B, a line BC and the angles ABC and BCA were measured, and found to be 322.55 yards, 60° 34', and 56° 10', respectively. Find the distance AB.

10. In making a survey by triangulation, the angles B and C of a triangle ABC were found to be 50° 30' and 122° 9', respectively, and the length BC is known to be 9 miles. Find AB and AC.

11. Two observers 5 miles apart on a plain, and facing each other, find that the angles of elevation of a balloon in the same vertical plane with themselves are 55° and 58°, respectively. Find the distance from the balloon to each observer, and also the height of the balloon above the plain.

12. In a parallelogram given a diagonal d and the angles x and y which this diagonal makes with the sides. Find the sides. Find the sides if d=11.237, x=19° 1', and y=42° 54'.

13. A lighthouse was observed from a ship to bear N. 34° E.; after sailing due south 3 miles, it bore N. 23° E. Find the distance from the lighthouse to the ship in both positions.

NOTE. The phrase to bear N. 34° E. means that the line of sight to the lighthouse is in the north-east quarter of the horizon, and makes, with a line due north, an angle of 34°.

14. In a trapezoid given the parallel sides a and b, and the angles x and y at the ends of one of the parallel sides. Find the non-parallel sides. Compute the results when a = 15, b=7, x=70°, y=40°.

Solve the following examples without using logarithms: 15. Given b=7.07107, A=30°, C=105°; find a and c. 16. Given c=9.562, A=45°, B=60°; find a and b. 17. The base of a triangle is 600 feet, and the angles at the base are 30° and 120°. Find the other sides and the altitude.

18. Two angles of a triangle are, the one 20°, the other 40°. Find the ratio of the opposite sides.

19. The angles of a triangle are as 5: 10: 21, and the side opposite the smallest angle is 3. Find the other sides.

20. Given one side of a triangle equal to 27, the adjacent angles equal each to 30°. Find the radius of the circum

scribed circle. (See § 33, Note.)

§ 38. CASE II.

Given two sides a and b, and the angle A opposite to the side a; find the remaining parts B, C, c.

This case, like the preceding case, is solved by means of the Law of Sines.

When an angle is determined by its sine it admits of two values, which are supplements of each other (§ 24); hence, either value of B may be taken unless excluded by the conditions of the problem.

If ab, then by Geometry A> B, and B must be acute whatever be the value of A; for a triangle can have only one obtuse angle. Hence, there is one, and only one, triangle that will satisfy the given conditions.

If a =

b, then by Geometry A=B; both A and B must be acute, and the required triangle is isosceles.

If ab, then by Geometry A<B, and A must be acute in order that the triangle

may be possible. If A is
acute, it is evident from
Fig. 34, where / BACA,
AC=b, CB = CB'
CB' = a,
that the two triangles ACB
and ACB' will satisfy the
given conditions, provided
a is greater than the per-

pendicular CP; that is, provided a is greater than b sin A (§ 11). The angles ABC and AB'C are supplementary (since ABC BB'C); they are in fact the supplementary angles obtained from the formula

If, however, a=b sin A= CP (Fig. 34), then sin B1, B=90°, and the triangle required is a right triangle.

If a <b sin A, that is, < CP, then sin B>1, and the triangle is impossible.