How do you solve the triangle given a=20, c=24, B=47?

1 Answer
Dec 16, 2016

Triangle is #a=20#, #b=17.924#, #c=24#, #A=54.7^o#, #B=47^o# and #C=78.3^o#.

Explanation:

Solving a triangle means identifying length of all the three sides as well as measures of all three angles. This is generally done using Law of sines, which is #a/sinA=b/sinB=c/sinC# and Law of cosines, according to which #b^2=a^2+c^2-2ac cosB#, #c^2=a^2+b^2-2abcosC# and #a^2=b^2+c^2-2bc cosA#

Here, we are given #a=20#, #c=24# and #B=47^o#.

We can use #b^2=20^2+24^2-2xx20xx24xxcos47^o#

= #400+576-960xx0.682=976-654.72=321.28#

Hence #b=sqrt321.28=17.924#

Now using ^^Law of sines**

#20/sinA=24/sinC=17.924/(sin47^o)=17.924/0.7314=24.5064#

Hence #sinA=20/24.5064=0.8161# and #A=54.7^o#

and #sinC=24/24.5064=0.9793# and #C=78.3^o#

Hence, triangle is #a=20#, #b=17.924#, #c=24#, #A=54.7^o#, #B=47^o# and #C=78.3^o#.