Question

In △ABC, m∠A=23°, a=10, and b=13. Find c to the nearest tenth.

Answer 1

Using Sine Rule

`\rightarrow (a)/(\sinA)=(b)/(\sin B)=(c)/(\sin C)\\\\ \rightarrow (10)/(\sin 23^(\circ))=(13)/(\sin B)\\\\ \sinB=(13 * 0.39)/(10)\\\\ \sin B=(5.07)/(10)\\\\ \sinB=0.507\\\\B=31^(\circ)`

Using Angle sum property of Triangle

∠A+∠B+∠C=180°

23°+31°+∠C=180°

∠C=180°-54°

∠C=126°

Again using Sine rule

`(c)/(\sin C)=(b)/(\sin B)\\\\ (c)/(\sin126^(\circ))=(13)/(0.507)\\\\c=(0.809 * 13)/(0.507)\\\\c=(10.517)/(0.507)\\\\c=20.7435`

Length of third side=20.75(approx)

Answer 2

`c=20.6`

Explanation:

We know that m∠A=23°, a=10, and b=13

To find c we must first find B and C. Then we use the sine theorem

sine theorem:

`(sin(A))/(a)=(sin(B))/(b) =(sin(C))/(c)`

then:

`(sin(A))/(a)=(sin(B))/(b)`

`(sin(23))/(10)=(sin(B))/(13)`

`0.0391=(sin(B))/(13)`

`Arcsin(13*0.0391)=B`

`B=30.55`

So

We know that:

`23+30.55+C=180\\C=180-23-30.55\\C=126.45`

we use the sine theorem again to find the length c

`(sin(23))/(10)=(sin(C))/(c)`

`0.0391=(sin(C))/(c)`

`c=(sin(126.45))/(0.0391)`

`c=20.6`