Plane trigonometry, by S.L. Loney.

PEDAL TRIANGLE.

237

We have PK = KB tan PBK = KB tan (90~ - C)
= ABcos B cot C = cos Bcos
sin C
= 2R cos B cos C (Art. 200).
Again AP = AK-PK= c sinB-PK
= 2R sin C sin B - 2R cos B cos C
=- 2R cos (B + C)
= 2R cos A (Art. 72).
So BP = 2R cos B, and CP = 2R cos C.
The distances of the orthocentre from the angular
points are therefore 2R cos A, 2R cos B and 2R cos C; its
distances from the sides are 21R cos B cos C, 2R cos C cos A,
and 21 cos A cos B.
210. To find the sides and angles of the pedal triangle.
Since the angles PKC and PLC are right angles, the
points P, L, C, and K lie on a circle..'. zPKL = Z PCL (Euc. III. 21)
= 90~-A.
Similarly P, K, B, and 1V lie on a circle, and therefore
z PKM= Z PBM
=90 -A.
Hence Z MKL = 180~- 2A
= the supplement of 2A.
So Z KLM= 180 - 2B,
and z LMK=180~-2C.