How do you solve the triangle given $A=24.3^circ, C=54.6^circ, c=2.68$ ?

Question

3897 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-29T12:15:57

Solution of triangle is $a ~~ 1.35, b ~~ 3.23, c=2.68$
and $/_A=24.3 , /_B=101.1 , /_C=54.6$

Angle between Sides $a and b$ is $/_C= 54.6^0$

Angle between Sides $b and c$ is $/_A=24.3^0 :.$

Angle between Sides $c and a$ is

$/_B=180-(54.6+24.3)=101.1^0$ The sine rule states if

$a, b and c$ are the lengths of the sides

and opposite angles are $A, B and C$ in a triangle, then:

$a/sinA = b/sinB=c/sinC ; c=2.68 ; b/sinB=c/sinC$ or

$b/sin101.1=2.68/sin54.6 or b= 2.68* (sin101.1/sin54.6)$ or

$b ~~ 3.23 (2dp)$ Similarly $a/sinA=c/sinC$ or

$a/sin24.3=2.68/sin54.6 or a= 2.68*(sin24.3/sin54.6)$ or

$a ~~ 1.35 (2dp) :.$ Solution: $a ~~ 1.35, b ~~ 3.23, c=2.68$

and $/_A=24.3 , /_B=101.1 , /_C=54.6$ [Ans]

How do you solve the triangle given A=24.3^circ, C=54.6^circ, c=2.68A=24.3^circ, C=54.6^circ, c=2.68?