Question

Solve the right triangle.a=1.9 in., A=46.5°, C=90°Round to one decimal place

Answer 1

So we have to solve a right triangle i.e. find all of its sides and angles. Assuming that angles are labeled with letters in capitals and that the opposite side of an angle is labeled with the same letter but in lower case then the triangle looks like this:

First of all it's important to remember that the sum of the internal angles of a triangle is equal to 180°. Then we have:

`\begin{gathered} 180^(\circ)=A+B+C \\ 180^(\circ)=46.5^(\circ)+B+90^(\circ) \\ 180^(\circ)=B+136.5^(\circ) \end{gathered}`

Then we substract 136.5° from both sides:

`\begin{gathered} 180^(\circ)=B+136.5^(\circ) \\ 180^(\circ)-136.5^(\circ)=B+136.5^(\circ)-136.5^(\circ) \\ 43.5^(\circ)=B \end{gathered}`

So we found angle B now we just need to find sides b and c. The law of sines states that the quotient between the length of a side and its opposite angle is the same for the three pairs of angles-opposite sides. Then we get:

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

With the first two terms we get:

`\begin{gathered} (a)/(\sin A)=(b)/(\sin B) \\ (1.9in)/(\sin 46.5^(\circ))=(b)/(\sin 43.5^(\circ)) \end{gathered}`

We can multiply both sides by the sine of 43.5°:

`\begin{gathered} (1.9in)/(\sin46.5^(\circ))=(b)/(\sin43.5^(\circ)) \\ (1.9in)/(\sin46.5^(\circ))\cdot\sin 43.5^(\circ)=(b)/(\sin43.5^(\circ))\cdot\sin 43.5^(\circ) \\ 1.8in=b \end{gathered}`

So b=1.8in. Then for c we can use the equality in the right side of the law of sines:

`\begin{gathered} (1.8in)/(\sin43.5^(\circ))=(c)/(\sin90^(\circ))=(c)/(1)=c \\ 2.6in=c \end{gathered}`

Then the answers are:

a=1.9 in

b=1.8 in

c=2.6 in

A=46.5°

B=43.5°

C=90°