A triangular parcel of a land with vertices at R,S, and T was to be enclosed by a fence, bit it was discovered that the surveyor's mark at S was Missing . From a deed to the property , it was learnedthat the distance from T to R is 325 m, the distance from T to S is 550m , and the angle at Rin triangle is 120°, determine the location of S by finding the distance from R to S.
Answers
Answer:
For this kind of problem we use the concept of trigonometry. You should remember:-
Cosine rule(for any triangle),
a^2=b^2+c^2-2bc.cosA.
OR, b^2=a^2+c^2-2ac.cosB.
OR,c^2=a^2+b^2-2ab.cosC.
And the SINE rule
sinA/a=sinB/b=sinC/c.
For this case we are given;
RT=324m
TS=506m
angle at R(R=125.4*).
We're required to find
angle S and RS.
Firstly we use cosine rule.
TS^2=RT^2+RS^2-2RT.RScosR.
(506)^2=(324)^2+RS^2-(2*234*RScos(125.4)).
On rearranging in the quadratic form we get,
RS^2-648cos(125.4)+(324^2-506^2)=0.
Plag in the calculator and you can find that RS=243.92m.
And angle S is;
sinS/324=sinR/506.
SinS=(324/506)sin(125.4)
sinS=0.5219.
S=sin-1(0.5219).
S=31.46¤. Therefore the distance from R and S is 243.92m and the angle at S is 31.46¤.
Step-by-step explanation:
For this kind of problem we use the concept of trigonometry. You should remember:-
Cosine rule(for any triangle),
a^2=b^2+c^2-2bc.cosA.
OR, b^2=a^2+c^2-2ac.cosB.
OR,c^2=a^2+b^2-2ab.cosC.
And the SINE rule
sinA/a=sinB/b=sinC/c.
For this case we are given;
RT=324m
TS=506m
angle at R(R=125.4*).
We're required to find
angle S and RS.
Firstly we use cosine rule.
TS^2=RT^2+RS^2-2RT.RScosR.
(506)^2=(324)^2+RS^2-(2*234*RScos(125.4)).
On rearranging in the quadratic form we get,
RS^2-648cos(125.4)+(324^2-506^2)=0.
Plag in the calculator and you can find that RS=243.92m.
And angle S is;
sinS/324=sinR/506.
SinS=(324/506)sin(125.4)
sinS=0.5219.
S=sin-1(0.5219).
S=31.46¤. Therefore the distance from R and S is 243.92m and the angle at S is 31.46¤.