(a)
![enter image source here]()
As DeltaABC is a right triangle, using Pythagorean theorem we get,
AC^2=AB^2+BC^2
=> AC=sqrt(2.5^2+3.5^2)=sqrt18.5=4.301m
Total length =AB+BC+AC
=2.5+3.5+4.301=10.301=10.3m (3 significant figures).
If AB is not part of the framed structures, then
Total length =BC+AC=3.5+4.301=7.801m=7.80m (3 significant figures)
tanangleA=(BC)/(AB)=3.5/2.5
=> angleA=tan^-1(3.5/2.5)=54.5^@
=> angleC=90-54.5=35.5^@
(b)
![enter image source here]()
AC^2=AB^2+BC^2
=> BC=sqrt((4.5^2-2.9^2))=3.441m
Total length =AB+BC+AC
=2.90+3.441+4.5=10.841=10.8m (3 significant figures).
If AB is not part of the framed structures, then
Total length =BC+AC=3.441+4.5=7.941m=7.94m (3 significant figures).
cosangleA=(AB)/(AC)=2.9/4.5
=> angleA=cos^-1(2.9/4.5)=49.9^@
=> angleC=90-49.9=40.1^@
(c)
![enter image source here]()
As DeltaABC is a right triangle and AB is the hypotenuse,
=> AC=ABcos50=2.0*cos50=1.286m
=> BC=ABsin50=2.0*sin50=1.532m
angleABC=90-50=40^@
angleCBD=90-40=50^@
angleBCD=180-50-20=110^@
In DeltaBCD, using sine rule we get,
(BC)/sin20=(BD)/sin110=(CD)/sin50
=> BD=1.532*sin110/sin20=4.209m
=> CD=1.532*sin50/sin20=3.431m
Total length =AB+BC+AC+BD+CD
=2.0+1.532+1.286+4.209+3.431
=12.458m=12.5m (3 significant figures)
If AB is not part of the framed structures, then
Total length =BC+AC+BD+CD
=1.532+1.286+4.209+3.431
=10.458m=10.5m (3 significant figures)
(d)
![enter image source here]()
BC=3/cos55=5.230
BD=3tan55=4.284
As ABCD is a rectangle,
=> AB=CD=3.0, and AC=BD=4.284
In DeltaBDE, angleBDE=90-38=52^@
angleDBE=180-52-25=103^@
=> (BD)/sin25=(BE)/sin52=(DE)/sin103
=> BE=4.284*sin52/sin25=7.988
=> DE=4.284*sin103/sin25=9.877
total length = 2(AB+BD)+BC+BE+DE
=2(3.0+4.284)+5.230+7.988+9.877
=37.663m=37.7m (3 significant figures)
If CD is not part of the framed structures, then,
total length = AB+2BD+BC+BE+DE
=3.0+2xx4.284+5.230+7.988+9.877
=34.663m=34.7m (3 significant figures)
