Question

find the missing parts of the triangle given below with 3 identified parts?

1.A=70 degrees 54' B=79 degrees 6' and a=20m

2.B=36 degrees 10' a=21m and b=30m

3.B=60degrees a=50m and c=60m

4.a=20m b=30m and c=40m

Answer 1

Solution:

1) $A=70^{0}54'=70.9^{0};$

$B=79^{0}6'=79.1^{0};$

$a=20m.$ Let's find angle $C:$

$C=180-79.1-70.9=30^{0};$

From the sine theorem find other sides of triangle.

$\frac{a}{sinB}=\frac{b}{sinC};$

$b=a\frac{sinC}{sinB}=20\times\frac{sin30}{sin79.1}=10.18m;$

$c=a\frac{sinA}{sinB}=20\times\frac{sin70.9}{sin79.1}=19.25m;$

2) $B=36^{0}10'=36.16^{0};$

$a=21m ; b=30m;$

$\frac{a}{sinB}=\frac{b}{sinC};$

$sinC=sinB\times(\frac{b}{a})=\frac{30}{21}\times sin36.16=0.84;$

$C=arcsin(0.84)=57.14^{0};$

$A=180-B-C=180-36.16-57.14=86.7^{0};$

$c=b\frac{sinA}{sinC}=30\times\frac{sin86.7}{sin57.14}=35.65m.$

3) $B=60^{0}; \space a=50m; \space c=60m;$

$sinC=\frac{a}{b}sinB=\frac{50}{60}sin60=0.72;$

$C=arcsin(0.72)=46.1^{0};$

$A=180-60-46.1^{0}=73.9;$

$b=c\frac{sinA}{sinB}=60\times\frac{sin73.9}{sin60}=66.56m.$

4) $a=20m;\space b=30m;\space c=40m;$

Let's use the cosine theorem:

$a^{2}=b^{2}+c^{2}-2bccosC;$

$cosC=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{900+1600-400}{2\times30\times40}=0.87;$

$C=arccos(0.87)=29.54^{0};$

From the sine theorem:

$sinA=\frac{b}{a}sinC=\frac{30}{20}\times sin29.54=0.74;$

$A=arcsin(0.74)=47.73^{0};$

$B=180-29.54-47.73=102.73^{0}.$

Answer:

1) $C=30^{0}; \space b=10.18m;\space c=19.25m.$

2)

$C=57.14^{0}; \space A=86.7^{0}; \space c=35.65m.$

3)

$A=73.9^{0}; \space C=46.1^{0};\space b=66.56m.$

4)

$A=47.73^{0}; \space B =102.72^{0};\space C=29.54^{0}.$