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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!

Iggy
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  • If you have been working on sinus recently, you should find the mesure of angle A in DAC. Angle A in BAC is easy to find. Then you can answer question a) with a substraction... – Evargalo Jun 09 '17 at 16:31
  • For question b), you can simlilarly find the area of DAC as the difference between the areas of ABC and DAC... – Evargalo Jun 09 '17 at 16:32
  • If AC = BC then the have are right iscoslese triangle. so what is angle CAB. If DB = AD/2 you have a right triangle with a hypotenuse of 2 units and a leg of 1 unit. So what is angle CAD. So what are all the other angles? What is the area of a right isoceles triangle. What is the area of a right triangle with a base half the hypotenuse? What do you get when you substract them. – fleablood Jun 09 '17 at 16:33
  • @Evargalo no need for trig. This a very basic right triangles. – fleablood Jun 09 '17 at 16:34
  • Basic right triangle. 1/2 and 1 and altitude is what type of triangle? (It is what type of triangle cut in half? What type of triangle with have sides of 1 and 1 and 1?) – fleablood Jun 09 '17 at 16:48

2 Answers2

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[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

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

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

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

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

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

I hope this answers your question!

Community
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CaptainSora
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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 $\triangle BAD$

fleablood
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  • Thank you for the explanation, but I still don't get how I can figure out CAB if I know that BCA = 90 – Iggy Jun 09 '17 at 17:09