Question

Determine the area of the triangle when tana=1 the image is in the picture above

Answer 1

The area will be:

`A=18(1+√(3)) \: u^(2)` or
`A=49.18 \: u^(2)`.

Explanation:

Using the definition of tangent, we have:

`tan(\alpha)=\frac{6}{\vec{AB}}`

We know that tan(α) = 1, then we can find AB

`1=\frac{6}{\vec{AB}}`

`\bar{AB}=6\: u`

u means any unit.

Now, we need to find the distance BC. If the angle ∠D is 60°, then ∠C must be 30°. Using the tangent definition in the triangle BCD we have:

`tan(30)=\frac{6}{\bar{BC}}`

`\bar{BC}=(6)/(tan(30))`

`\bar{BC}=(6)/(1/√(3))`

`\bar{BC}=6√(3) \: u`

So, the base of the triangle will be:

`b=\bar{AB}+\bar{BC}=6+6√(3)=6(1+√(3)) \: u`

The area of a triangle is given by the following equation:

`A=(b*h)/(2)`

• b is the base (
  `b=6(1+√(3))\: u`)
• h is the height (h=6 u)

`A=(6(1+√(3))*6)/(2)`

`A=18(1+√(3))`

Therefore, the area will be
`A=49.18 \: u^(2)`.

I hope it helps you!