Question

Triangle ΔABC has side lengths of a = 16, b equals 16 times radical 3 comma and c = 32 inches. Part A: Determine the measure of angle A period (5 points) Part B: Show how to use the unit circle to find tan A. (2 points) Part C: Calculate the area of ΔABC. (3 points)

Answer 1

A=30

TanA =
`(√(3) )/(3)\\`

Area = 221.7025034
`in^(2)`

A) Using cosine rule,

`a^(2) = b^(2) + c^(2) - 2bccosA`

`16^(2) = (16√(3) )^(3) + 32^(2) - 2(16√(3) )32cosA`

A=30

B) cosA =
`(√(3) )/(2)`

From the unit circle, we search for Tan A in the first quadrant since cos A is in the first quadrant.

TanA =
`(y)/(x)` =
`((1)/(2) )/((√(3) )/(2))` =
`(√(3) )/(3)`

C) Area =
`(1)/(2) bcsinA` =221.7025034
`in^(2)`

Answer 2

Step 1

Given;

Step 2

`A)\text{ Determine m}\angle A`

Using cosine rule;

`\begin{gathered} a^2=b^2+c^2-2bccosA \\ 16^2=(16√(3))^3+32^2-2(32)(16√(3))cosA \end{gathered}`
`256=768+1024-1773.620037cos\text{ A}`
`\begin{gathered} -1536=-1773.620037cos\text{ A} \\ A=30.00000056^o \end{gathered}`

B) Below is a unit circle image

`\begin{gathered} At\text{ a point in our solution we had;} \\ -1536=-2(32)(16√(3))cosA \\ cosA=(√(3))/(2) \end{gathered}`

From the unit circle, we search for Tan A in the first quadrant since cos A is in the first quadrant.

`\begin{gathered} TanA=(y)/(x) \\ y=(1)/(2),x=(√(3))/(2) \\ TanA=((1)/(2))/((√(3))/(2)) \\ TanA=(1)/(√(3))*(√(3))/(√(3))=(√(3))/(3) \\ TanA=(√(3))/(3) \end{gathered}`

C) Calculate the area of triangle ABC

Using

`\begin{gathered} Area=(1)/(2)bcsinA \\ Area=(1)/(2)*16√(3)*32* sin(30) \\ Area=221.7025034in^2 \end{gathered}`

Answer;

`\begin{gathered} A)m\angle A=30^o \\ B)TanA\text{ =}(√(3))/(3) \\ C)AreaoftriangleABC=221.7025034in^2 \end{gathered}`