Solve each of the following triangles (if possible). Then, find the area of each. ?=32.7°, a=37.5...

Solution

1) B=32.7°, a=37.5cm, b=28.6cm

this is a special case so, first

use law of sine to find sinA

sinA/a=sinB/b

sinA=a×sinB/b=37.5×sin32.7°/28.6=0.7384

sinA<1 therefore, two triangles are possible

And A=arcsin(0.7384)=47.6°

and also, sin is positive in second quadrant

so, A=180°-47.6°=132.4°

now, solve first triangle with A=47.6°

By angle sum property of triangle

C=180°-A-B=180°-47.6°-32.7°=99.7°

use law of sine

sinC/c=sinB/b

c=b×sinC/sinB=28.6×sin99.7°/sin32.7°=52.2

now, solve second triangle with A=132.4°

by angle sum property of triangle

C=180°-132.4°-32.7°=14.9°

use law of sine

c=b×sinC/sinB=28.6×sin14.9°/sin32.7°=13.6

so, other values of triangle 1 are

A=47.6°, C=99.7° and c=52.2cm

values of second triangle are

A=132.4°, C=14.9° and c=13.6cm

to find the area use heron's formula

s=(a+b+c)/2 , A=√[s(s-a)(s-b)(s-c)]

here, a=37.5, b=28.6 and c=52.2 (for first triangle)

s=(37.5+28.6+52.2)/2=59.2

s-a=59.2-37.5=21.7

s-b=59.2-28.6=30.6

s-c=59.2-52.2=7

so, area=A=√[59.2×21.7×30.6×7]=√275169.88=524.6 (cm)^2

for second triangle

a=37.5 , b=28.6 and c=13.6

s=(37.5+28.6+13.6)/2=39.9

s-a=39.9-37.5=2.4

s-b=39.9-28.6=11.3

s-c=39.9-13.6=26.3

so, area=A=√[39.6×2.4×11.3×26.3]=√28423.3=168.6 cm^2

b)a=4.1in, b=9.8in, c=6.2in

first of all area by heron's formula

s=(4.1+9.8+6.2)/2=10.1

s-a=10.1-4.1=6

s-b=10.1-9.8=0.3

s-c=10.1-6.2=3.9

A=√[10.1×6×0.3×3.9]=√70.551=8.4 in^2

now, by law of cosine

a^2=b^2+c^2-2bc×cosA

cos A=-(a^2-b^2-c^2)/2bc

cosA=-[(4.1)^2-(9.8)^2-(6.2)^2]/2×9.8×6.2

cosA=117.7/121.52=0.9685

A=arccos(0.9685)=14.4°

now, usw law of sine

sinA/a=sinB/b

sinB=b×sinA/a=9.8×sin14.4°/4.1=0.5944

B=arcsin(0.5944)=36.5°

use angle sum property of triangle

C=180°-14.4°-36.5°=129.1°

so, other values are

A=14.4°, B=36.5°, C=129.1° and Area=8.4 in^2