Therefore or and abc sin a: sin ß: sin y. Q.E.D. Def. The common value of the three quotients sin y is called the modulus of the triangle. THEOREM II. The modulus of the triangle is equal to the diameter of the circumscribed circle. This theorem may be demonstrated by the student from the property that an inscribed angle is measured by one half the arc on which it stands. Theorem I. enables us, when the angles and one side of any plane triangle are given, to find the remaining two sides. If the side given is c, we have c sin a If we put p, p', and p" for the lengths of the perpendiculars from a, B, and y respectively upon the opposite sides, we find, from the preceding figure, By these equations we may find the lengths of the perpendiculars. 2. 3. 66 796.25; find b and c. a = 5° 26'.2, B ❝b and c. a = 50° 58'.7, 6 = 32° 50'.8, c = 169.37; "a and b. 4. In order to from the points A find the distance of a point across a river Required the distances AC and BC across the river, and the length of the perpendicular from Con AB. 5. In a triangle ABC the angle B exceeds the angle A by 10°, the angle C exceeds the angle B by 20°, and the side AC is 2.72 904 metres. Find the angles and sides of the triangle and the length of the perpendiculars from the angles upon the opposite sides. 6. The base of an isosceles triangle is 132.643 metres, and the angle at the vertex is 32° 53'.7. Find the sides and the altitude. 7. One diagonal of a parallelogram measures 23 metres, and it makes angles of 32° 17′ and 63° 24' with the sides. Find the lengths of the sides and the angles of the parallelogram. 8. From a point at a distance a from the centre of a circle of radius tangents are drawn to the circle. Express the lengths of the tangents and the distance between the points of tangency, and compute the result when r = 7, a = 12. 57. CASE II. Given two sides and the angle opposite one of them. Let the given parts be a, b, a. We then compute the parts ß, y, and c by the formulæ (a) and (3) already found. This case may have two solutions, as is shown in geometry. The two solutions are found in the above equations, because to a given value of sin 6 corresponds either of two angles (821), which will be supplements of each other. But if one solution gives a+B> 180° it is not admissible, and only the lesser value is used to give the triangle. This will be the case when a >b. It may also happen that sin ß = b a sin a comes out greater than unity. There is then no possible triangle which fulfils the con ditions. EXAMPLE. Given a = 152.08, b= 236.74, a= 32° 29'.6; find 6. From a point P at a distance a from the centre of a circle of radius a line is drawn, making the angle ẞ with the line from P to the centre of the circle. At what distances from P will the line intersect the circle? Compute the distances when r = 72, a 98, and ẞ= 28° 56'. 7. Show that if, in the present case, we take the side which is not given as the base, we can find the altitude of the triangle immediately, and afterwards may find the three required parts of the triangle from the altitude. CASE III. Given the three sides. THEOREM III. In a 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 two sides into the cosine of the angle included by them. In symbolic language this theorem is expressed in any of the forms or or a2 = b2 + c2 2bc cos a, b2 = ac2 2ac cos B, c2 = a + b2 2ab cos y. (9) Proof. It is shown in geometry that in any triangle the side opposite an acute angle is greater than the sum of the squares of the other two sides by twice the product of one of these sides into the projection of the other side upon it. If a be the acute angle, we have, by this theorem, a2 = b2 + c2 26 × (projection of c on b). By $35, II., and the definition of projections. Projection of c on b = c cos a; substituting this value of the projection, a2 = b2 + c2 — 2bc cos a. Q.E.D. The other equations may be proved in exactly the same way. If the containing angle is obtuse, the square of the opposite side will be greater than the sum of the squares of the containing sides. But this case is included in the trigonometric formula, because then cos a is negative and be cos a is positive. Hence the formula is applicable to all cases when regard is had to the algebraic sign. which with the two companion formulæ enable us to find the angles when the three sides are given. 59. If the angle is small, it cannot be accurately determined by means of its cosine; we therefore transform the expression as follows: Subtracting each member of the equation from unity, we have 2bc — b3 — c2 + a2 1 cos a = 1 b2 + c2 — a2 2bc Let us now put s for half the sum of the three sides, so that Then s = {(a+b+c). a b c = 28 - 2c; ab+c=28-26; and the preceding equation reduces to (8 — b) (8 — 0) sin'α = bc (11) The expressions for the other two angles are obtained by the same process, the letters a, b, c and a, B, y being permuted in the orders b, c, a; ß, y, a and c, a, b; y, a, ß. We thus find These equations answer our purpose, but in determining an angle the tangent is the function to be preferred, because an angle can be determined more accurately from its tangent than from its sine or cosine. To obtain expressions for the tangent add unity to both sides of the equation (10). We then have Since 1+ cos a = 2cos2 ja (§44), this equation reduces to (b + c)2 — a2 2 cos2 a = 2bc (b+c+a) (b+c− a) 2bc |