Calculate the angle in the triangle

Question

Currently stuck at this one:

Triangle picture

In the figure, AC = BC and DC = 0.5*AD.

a) Calculate the angle A in the triangle BDA

b) Create an appropriate for the side AC in meters(Chose an appropriate size for the side AC) and calculate the area of the triangle BDA. Give different suggestions about how the triangle BDAs area can be decided.

I have no idea how to get any values out of this, could anyone please help me with this one? Thank you in advance!

Answer 1

Theorem: The base angles of an isoceles triangle are congruent.

THEREFORE:

1) The base angles of an isoceles right triangle of $45$ degrees.

Because: If then angles of an isoceles right triangle are $m = 90$ and $j=k$. Then $90 + j + k = 90 + 2k = 180$ and so $j=k = 45$.

2) The angles of an equilateral triangle are $60$ degrees.

Because: If two sides are equal the base angles are equal. If all sides are equal then all pairs of base angles are equal. So if the angles are $m = j=k$ then $m+j+k = 3m = 180$ so $m = j= k= 60$.

3) If a right triangle is such that one side is $\frac 12$ unit, then hypotenuse is $1$ units, than a) the third side is $\frac{\sqrt{3}}2$ units and b) the angles are $30$ (across from the $\frac 12$ unit side) and $60$ (across from the $\frac {\sqrt{3}}2$ side.

Because: If you cut an equilateral triangle in half you get a right triangle with angles $30$ and $60$ and $90$. The sides are the hypotenuse is one of the original sides. The base is half of another side. The third side is the altitude $a$ and $(\frac 12)^2 + a^2 = 1^2$. So $a = \frac {\sqrt{3}}2$.

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So that is enough to do everything.

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$AC = BC$ and $\triangle ABC$ is isoceles. $\angle BCA = 90$. So you can figure out $\angle CAB$.

$DC = AD/2$ so $AD = 2DC$ so $\triangle ADC$ is a right triangle with leg of length $DC$ and hypotenuse of $2DC$. SO you can figure out $\angle CAD$.

So you can figure out all the other angles.

You know $AC$ and $BC$ so you can find the area of $\triangle ABC$. You know $AC$ and $DC$ so you can find the area of $\triangle CAD$. Subtract them. That'll give you the area of triangle $\triangle BAD$

Answer 2

[Edit] Thanks fleablood for catching my error!

Part A

$\angle BDA = \angle BAC - \angle DAC$

$\sin(\angle DAC) = \frac{DC}{DA} = 0.5$

$\therefore \angle DAC = 30^\circ$

$\tan(\angle BAC) = \frac{BC}{AC} = 1$

$\therefore \angle BAC = 45^\circ$

$\therefore \angle BDA = 45^\circ - 30^\circ = 15^\circ$

Part B

Let $AC = 1m$

Method 1

$A = \frac{1}{2}bh$

$\because BD\bot AC$

$\therefore AC$ is a height to $BD$

$AD = \frac{AC}{\cos{30}} = \frac{2\sqrt{3}}{3}$

$A_{\triangle BDA} = \frac{1}{2}(BD)(AC)$

$A_{\triangle BDA} = \frac{1}{2}(BC-DC)(AC)$

$A_{\triangle BDA} = \frac{1}{2}(1m-\frac{\sqrt{3}}{3}m)(1m)$

$A_{\triangle BDA} = \frac{3-\sqrt{3}}{6}m^2$

Method 2

$A_{\triangle BDA} = A_{\triangle BCA} - A_{\triangle DCA}$

$A_{\triangle BDA} = \frac{1}{2}(AC)(BC) - \frac{1}{2}(AC)(DC)$

$A_{\triangle BDA} = \frac{1}{2}(1m)(1m) - \frac{1}{2}(1m)(\frac{\sqrt{3}}{3}m)$

$A_{\triangle BDA} = \frac{1}{2} - \frac{\sqrt{3}}{6}$

$A_{\triangle BDA} = \frac{3-\sqrt{3}}{6}m^2$

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I hope this answers your question!