Thus the ratio of the error to the estimated height is

(1) A triangle is solved from the given parts A, b, c; if there is a small error & in the angle A prove that the consequent error in the calculated area of the triangle B is approximately

8bc cos A.

(2) A triangle is solved from the given parts A, b, c; if there is a small error & radians in A, prove that the consequent error in B is - 8 sin B cos Ccosec A radians.

(3) If the sides of a triangle be measured and a small error c'exist in the measured value of c, prove that the consequent error in the diameter of the circumscribing circle is

c'cos A cos B

sin A sin B sin C

(4) The height and distance of an inaccessible object are found by observing the angles of elevation a and ß at two points A and B in a horizontal line through the base of the object, the distance between A and B being known; if the same error be made in each in consequence of an imperfect observation of the horizontal, show that the ratio of the error in the calculated height of the object to the calculated distance is

tan (a+B): 1.

(5) The area of a quadrilateral AOBQ right-angled at A and B is to be determined from observations of the angle AOB, and the length (p and q) of OA and OB. Prove that the area is

and that if a small error & be made in the observation of the angle AOB the consequent error in the area is

8. AB2. cosec2 AOB.

(6) If the angles of a triangle, as computed from slightly erroneous measurements of the length of its sides, be A, B, C, prove that approximately, a ẞy being the errors of the lengths, the consequent errors in the cotangents of the angles are proportional to

γ

B cos C+ y cos B-a, y cos A+ a cos C-B, a cos B+ß cos A −y divided respectively by sin A, sin B, sin C.

(7) It is observed that the altitude of the top of a mountain at each of the points AB and C where ABC is a horizontal triangle is a. Shew that the height of the mountain is

a tan a cosec A.

If there be a small error n" in the altitude at C the true height is very nearly

(8) If in a triangle ABC the observed lengths of a, b, c are 5, 4, 6 and there is known to be a small error in the measurement of c, determine which angle can be found from the formula

CHAPTER X.

EXAMPLES OF THE APPLICATION OF TRIGONOMETRY TO GEOMETRICAL PROBLEMS.

121. In this chapter we shall use the following notation:

D, E, F are the feet of the perpendiculars drawn from the angular points A, B, C of the triangle ABC to the opposite sides.

AD, BE, CF intersect in a point P which is called the orthocentre of the triangle ABC.

DEF is called the pedal triangle of the triangle ABC. A'B'C' are the middle points of the sides BC, CA, AB. AA', BB', CC' intersect in a point G, which is called the centre of gravity of the triangle ABC.

3

I, I, I, I, are the centres of the inscribed and escribed circles of the triangle ABC; r, r1, r2, r, are their radii.

3

[E. 276, 278.] O is the centre of the circumscribing circle and R its radius.

The circumscribing circle of the triangle DEF passes through A'B'C' and through the middle points of each of the lines PA, PB, PC. It is called the nine-points circle. We shall denote its centre by N.

[Proofs of the propositions referred to above may be found in the appendix to Todhunter's Euclid.]

L.

EXAMPLE 1. To prove that P, N, G and O are in one straight line; and that PG=2GO=4NG, i.e. that N is the point of bisection, and G a point of trisection of PO.

N is the centre of a circle passing through D and A'. Therefore N lies in the line bisecting DA' at right angles. This line produced bisects OP. Again, the nine-points circle passes through E and B', therefore its centre N lies on the line bisecting EB' at right angles. This line produced also bisects OP. Therefore N is the middle point

Hence if AA' cut PO in G, AG: GA'=PA: OA'=2:1;

.. AG=2GA' or G is the centre of Gravity.

Also

PG: GO PA: OA'=2:1. Q.E.D.

122. It is often convenient, in attempting the solution of a geometrical problem, to express the lengths of lines involved each in terms of some common unit. When the problem is one concerning a triangle, the Radius of the Circumscribing circle may be employed as the unit. Its convenience is shewn by the symmetry of the following results:

EXAMPLES. XXXV.

Prove the following statements :

(1) a=2R sin A, b=2R sin B, c=2R sin C.

(2) s=R (sin A+ sin B+ sin C)=4R cos A. cos B. cos C.

(3) r=4R sin A. sin †B. sin ≥C.

(4) r1=4R sin A. cos B. cos C.

(10) The radius of the nine-points circle=R.

(11) The sides of the triangle DEF are R sin 24, R sin 2B, R sin 2C.

(12) The area of DEF=R2 sin 24. sin 2B. sin 2C.

(13) BD=R (2 cos B sin C+ sin ▲).

(14) dN=R cos (B – C').

(15) The distances of the centres of the escribed circles from that of the inscribed circles are

4R sin 4, 4R sin B, 4R sin C.

(16) AE'=R (sin B + sin C − sin A), where E' is the point in which the inscribed circle touches AC.