In Exercises 29-36, all references are to the figure, the distance between A and
ID: 3025894 • Letter: I
Question
In Exercises 29-36, all references are to the figure, the distance between A and D, is a, and the distance between D and C is x. Assuming that A = 37 degree, BDC = 49 degree, and a = 17, find h and x. Assuming that A = 21 degree, BDC = 53 degree, and a = 8, find h and x. Assuming that A = 27 degree, BDC = 54 degree, and c_1 = 33, find h, x, and a. Assuming that A = 14 degree, BDC = 31 degree, and c_2 = 50 find h, x, and a. Assuming that A = 39 degree, BDC = 55 degree, and c_2 = 22, find h, x, and a. Assuming that A = 28 degree, BDC = 36 degree, and c_2 = 19, find h, x, and a. Assuming that A = 13 degree, DBC = 25 degree, and a = 11, find h and x. Assuming that A = 24 degree, DBC = 31 degree, and a = 40. find h and x.Explanation / Answer
29) we are assuming <A=370,<BDC=490 and a=17
we get from given triangle is tan(A)=h/(a+x) since tan(theta)=opposite side/adjacent side
tan(A)=h/(a+x)=h/(17+x)
h=(17+x)tan(370) --(1) since <A=370 and a=17
tan(D)=h/x implies h=xtan(490) since <D=490
substitute h=xtan(490) into equation (1)
xtan(490) =(17+x)tan(370)
x(tan(490)-tan(370))=17tan(370)
x=(17*tan(370))/(tan(490)-tan(370))
x=32.28
substitute x into h=xtan(490)
h=32.28*tan(490)=37.13
30) we are assuming <A=210,<BDC=530 and a=8
we get from given triangle is tan(A)=h/(a+x) since tan(theta)=opposite side/adjacent side
tan(A)=h/(a+x)=h/(8+x)
h=(8+x)tan(210) --(1) since a=8 and <A=210
tan(D)=h/x implies h=xtan(530) since <BDC=530
substitute h=xtan(530) into equation (1)
xtan(530)=(8+x)tan(210) since a=8
x(tan(530)-tan(210))=8tan(210)
x=(8tan(210))/(tan(530)-tan(210))=3.25
x=3.25
substitute x into h=xtan(530)
h=3.25*tan(530)=4.32
h=4.32
31) we are assuming <A=270,<BDC=540 and c1=33
from given triangle we get tan(A)=h/(a+x) ,tan(D)=h/x and sin(A)=h/c1 implies sin(270)=h/33 implies h=33*sin(270)=14.98
substitute h=14.98 into tan(D)=h/x
tan(540)=14.98/x implies x=14.98/tan(540)=10.88
substitute h,x into tan(A)=h/(a+x)
tan(270)=14.98/(a+10.88) since h=14.98 and x=10.88
a+10.88=14.98/tan(270)
a=(14.98/tan(270))-10.88=18.51
32) we are assuming <A=140,<BDC=310 and c1=50
from given triangle we get tan(A)=h/(a+x) ,tan(D)=h/x and sin(A)=h/c1 implies sin(140)=h/50 implies h=50*sin(140)=12.09
substitute h=12.09 into tan(D)=h/x
tan(310)=12.09/x since h=12.09 and <BDC=310
x=12.09/tan(310)=20.12
substitute x,h into tan(A)=h/(a+x)
tan(140)=12.09/(a+20.12) since <A=140 ,x=20.12 and h=12.09
(a+20.12)=12.09/tan(140)
a=(12.09/tan(140))-20.12=28.37
33) we are assuming <A=390,<BDC=550 and c2=22
from given triangle we get tan(A)=h/(a+x) ,tan(D)=h/x
and sin(D)=h/c2 implies sin(550)=h/22 implies h=22*sin(550)=18.02
substitute h into tan(D)=h/x
tan(550)=18.02/x since h=18.02 and <BDC=550
x=18.02/tan(550)=12.62
substitute x,h into tan(A)=h/(a+x)
tan(390)=18.02/(a+12.62) since <A=390,x=12.62 and h=18.02
a+12.62=18.02/tan(390)
a=(18.02/tan(390))-12.62=9.63
34)we are assuming <A=280,<BDC=360 and c2=19
from given triangle we get tan(A)=h/(a+x) ,tan(D)=h/x
and sin(D)=h/c2 implies sin(360)=h/19 implies h=19*sin(360)=11.17
substitute h into tan(D)=h/x
tan(360)=11.17/x since h=11.17 and <BDC=360
x=11.17*tan(360)=8.12
substitute x,h into tan(A)=h/(a+x)
tan(280)=11.17/(a+8.12) since <A=280 ,h=11.17 and x=8.12
(a+8.12)=11.17/tan(280)
a=(11.17/tan(280))-8.12
a=12.88
35) we are assuming <A=130,<DBC=250 and a=11
from given triangle we get tan(A)=h/(a+x) ,tan(B)=x/h implies x=h*tan(B)
substitute x into tan(A)=h/(a+x)
tan(130)=h/(11+h*tan(250)) since x=h*tan(250) and a=11
(11tan(130)+h*tan(250)tan(130))=h
h(1-tan(250)tan(130))=11*tan(130)
h=(11*tan(130))/(1-tan(250)tan(130))=2.84
substitute h into tan(B)=x/h
x=2.84*tan(250)=1.33
36)we are assuming <A=240,<DBC=310 and a=40
from given triangle we get tan(A)=h/(a+x) ,tan(B)=x/h implies x=h*tan(B)
substitute x into tan(A)=h/(a+x)
tan(240)=h/(40+h*tan(310)) since <A=240 ,<B=310 and a=40
40*tan(240)+h*tan(310)tan(240)=h
h(1-tan(310)tan(240))=40*tan(240)
h=(40*tan(240))/(1-tan(310)tan(240))=24.32
substitute h into x=h*tan(B)
x=24.32*tan(310)=14.62 since h=24.32