\fdO'^ - Old Forge Coal Mines

158 GEOMETRY AND TRIGONOMETRY.
{d) See Fig. 13. Angle B A C=79° 22'; angle A B D
= 90° - 79° 22' = 10° 38'. Side A B-B D B
-=- sin 79° 22' = 12 ^ .98283 = 12.209 inches.
Side A D = B Z> X tan 10° 38' = 12 X
.18775 = 2.253 inches.
Side D C=A C-A Z>= 9.5 - 2.253 = a^
7.247 inches. fig. is
Side B C= VWW^^^WT' = 4/12' + 7.247-^ = 4/196.519 =
14.018 inches.
Perimeter of triangle equals A B-{- B C -\- A C = 12.209
+ 14. 018 + 9.5 = 35. 73 inches. Ans.
(333) The diagonal divides the trapezium into two
triangles; the sum of the areas of these two triangles equals
the area of the trapezium, which is, therefore,
11X7 . llx4i , .
—-—• , ^.7 • +
^ —~ = Gl- square mches. Ans.
li li O
(334) Referring to Fig. 17, problem 350, we have OA
10 3
or O B = — or 5 inches, and A B = 6- inches.
2 4
Sin C O ^=-§^ = ^^1"^ =. 675 ; therefore, angle COB =
42° 27' 14.3'.
Angle AOB= (42° 27' 14.3") X 2 = 84° 54' 28.6".
C0= OB Xcos COB = 5 X .73782=3.6891.
Ans.
Area of sector = 10' X .7854 X ^^
'
onJ^'^ = 18.524 sq.in
ooO
. .
^
6.75 , X 3.6891
Area ,^ ,.^
of triangle = = 12.4o0 sq. m.
18.524-12.450 = 6.074 sq. in., the area of the segment.
Ans.
(335) Convex area =
= r = 535. 5 square
perimeter of base X slant height 63 X 17 „
-r ^—
inches. Ans.
/i At