Chapter 5.8B, Problem 2GCE

To determine

To use SOLVTRI to solve the triangles in example 2b of lesson 5-7.

Explanation of Solution

Given:

Refer to example 2b.

The given set of values are −

  $a=24,b=40,c=18$

Calculation:

The following is the program of SOLVTRI

PROGRAM: SOLVTRI

: Disp “INPUT VERTEX A”

: Input A: Input P

: Disp “INPUT VERTEX B”

: Input B: Input Q

: Disp “INPUT VERTEX C”

: Input C: Input S

: $\sqrt{{\left(A -B\right)}^2+{\left(P -Q\right)}^2}\to E$

: $\sqrt{{\left(B-C\right)}^2+{\left( Q-S\right)}^2}\to F$

: $\sqrt{{\left(A -C\right)}^2+{\left(P -S\right)}^2}\to G$

  $\begin{array}{l}:co{s}^{-1}\left(\left(F^2- E^2 -G^2\right)/\left(-2EG\right)\right)\to M\\ :co{s}^{-1}\left(\left(G^2- E^2 -F^2\right)/\left(-2EF\right)\right)\to N\\ :co{s}^{-1}\left(\left(E^2- F^2 -G^2\right)/\left(-2FG\right)\right)\to O\end{array}$

: Disp “SIDE AB” :Disp E

: Disp “SIDE BC” :Disp F

: Disp “SIDE AC” :Disp G

: Pause

: Disp “ANGLE A” :Disp M

: Disp “ANGLE B” :Disp N

: Disp “ANGLE C” :Disp O

: Stop

INPUT:

  $a=24,b=40,c=18$

OUTPUT:

SIDE AB: $24$

SIDE BC: $40$

SIDE CD: $18$

ANGLE A: ${20.6}^{\circ }$

ANGLE B: ${144.1}^{\circ }$

ANGLE C: ${15.3}^{\circ }$