OBLIQUE TRIANGLES

460. THEOREM. In any triangle the three sides are proportional to the sines of the opposite angles.

In like manner by drawing the altitude on the side
BC it can be proved that

461. The law of sines. The preceding theorem is called the law of sines. By means of this law any triangle may be solved, (1) when two angles and a side are known; and (2) when two sides and an angle opposite one of them are known.

NOTE. In case (2) two solutions are possible. This results from the fact that the angles are computed from the sine, and supplementary angles have the same sine, § 457.

1. Solve the triangle ABC, given a = 50.0, A 10° 12', B = 46°, 36'.

Ans. C 123° 12', b =

=

2. Solve the triangle ABC, given c 5.0, A :

=

205.1, c = 236.4. = 58°, C = 57°.

463. THEOREM. In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these sides and the cosine of the included angle.

Proof. 1. Draw the altitude BD.

2. Then in the right ▲ ABD, c2 = BD2 + AD2. Why?

=

=

· a2 (sin C)2 + b2 −2 ab cos C + a2 (cos C)2

= a2{(sin C)2 + (cos C)2} + b2 −2 ab cos C.

5. ... c2 = a2 + b2 - 2 ab cos C.

§ 458

In like manner by drawing the altitudes on the

other two sides, it can be proved that

a2 = b2 + c2 — 2 bc cos A; b2 = c2 + a2 − 2 ac cos B.

464. The law of cosines. The theorem of § 463 is called the law of cosines. By means of this law the angles of a triangle may be computed when the three sides are known. For,

Exercise.
SUGGESTION: Let a = 7, b = 9, c = 11. Cos C =

If the sides of a triangle are 7, 9, 11, find the greatest angle.
a2 + b2 - c2
§ 464.

2 ab

COMPUTATION PROBLEMS

465. PROBLEM.

three sides, a, b, c.

To find the altitude of a triangle, given the

466. PROBLEM. Given a, b, c, the sides of a triangle, to find the area.

SOLUTION: 1. Draw the altitude he, and denote the area by S.

467. PROBLEM.

Given a, b, c, the sides

of a triangle, to find sin A, sin B, and sin C.

SOLUTION: 1. Sin A

=

hc
b

2. Hence sin A

1. If two sides of a triangle are each equal to 8 and the included angle is 60°, find the third side.

SUGGESTION: Apply the law of cosines, § 463.

2. If two sides of a triangle are each equal to 10 and the included angle is 120°, find the third side.

3. If the sides of a triangle are 7, 9, 12, find the greatest angle.

4. If the sides of a triangle are 6, 7, 9, find the three altitudes.

5. If the sides of a triangle are 3 ft., 4 ft., and 6 ft., find the area, and also the three angles.

SUGGESTION: Apply §§ 464 and 466.

6. If b and c are two sides of a triangle and A is the included angle, show that its area is bc sin A.

SUGGESTION: heb sin A.

7. Find the areas of the triangles in Ex. 1 and Ex. 2 by Ex. 7.

√3.

8. If a is the side of an equilateral triangle, show that its area is 9. AB = 500 ft. is a base line along the bank of a river, and C is an object on the opposite bank. If / BAC 57°, and Z CBA = 80° 20′, find the breadth of the river.

=

SUGGESTION: Apply the law of sines.

469. PROBLEM. Given a, b, c, the sides of a triangle, to find the radius R of the circumscribed circle.

SOLUTION: 1. Draw the diameter CD and join BD.

2. Then ▲ CDB is a rt. ▲ and CD is the hypotenuse. Why?

A=▲ CDB. Why?

C

b

A

Why?

abc

... a CD sin ▲ CDB = 2 R sin A.

3. Also

4.

5.

4 S
bc 4 S

(solving for R and substituting the value of

B

470. PROBLEM.

FIG. 212.

Given a, b, c, the sides of a triangle, to find

1. If the sides of a triangle are 5 ft., 12 ft., and 13 ft., find the area of the triangle; and also the radii of the circumscribed and inscribed circles. 2. If the sides of a triangle are 5, 5, 8, find the area, the angles, and the radii of the circumscribed and the inscribed circles.

3. Given a, b, two adjacent sides of a parallelogram, and A, the included angle. Show that the area of the parallelogram is ab sin A.

SUGGESTION: Find the altitude on one side.

4. If two sides of a parallelogram are 4 ft. and 6 ft., and the included angle is 60°, find the diagonals and the area. Apply the law of cosines.

5. A ladder 40 feet long reaches a window 33 feet high, on one side of a street. When turned over upon its foot, it reaches another window 21 feet high, on the opposite side of the street. Find the width of the street.