Triangles!

**ADY** · August 24th 2009, 10:20 AM

If you have a triangle

AC = 8cm
AB = 3cm
BC = no length given but C = 20° (Base)

Consider that triangle ABC. Now there is a further line AD ( D is on the base line and is a right angled) AD is perpendicular to BC and B lies between D and C.

How do i find the length of AD?

Thank you

**running-gag** · August 24th 2009, 10:45 AM

Hi

If I have well understood the situation is the following

In this case $\sin \alpha = \frac{AD}{AC}$ therefore $AD = AC \: \sin \alpha$

**ADY** · August 24th 2009, 10:54 AM

Well understood!

I make the length AD , 2.74cm is that right?

**ADY** · August 24th 2009, 11:14 AM

Now, how do i find the angle of ABD and ABC?

**masters** · August 24th 2009, 11:45 AM

Originally Posted by ADY

Well understood!

I make the length AD , 2.74cm is that right?

Correct.

Originally Posted by ADY

Now, how do i find the angle of ABD and ABC?

$\sin \angle ABD = \frac{2.74}{3}$ , so find $\arcsin \frac{2.74}{3}$ to find the measure of $\angle ABD$ .

$\angle ABC$ is supplementary to $\angle ABD$ .

**ADY** · August 24th 2009, 12:27 PM

so how do inverse sine? $sin^-1 \frac{2.74}{3} = 1.2$ ?

**mathaddict** · August 24th 2009, 07:21 PM

Originally Posted by ADY

so how do inverse sine? $sin^-1 \frac{2.74}{3} = 1.2$ ?

How did you get 1.2 by the way .

to find the inverse sine, from your calculator , press shift , sin , then you will see sin^{-1} appearing on the screen , then you key in (2.74/3) to get 65.97.

**ADY** · August 24th 2009, 10:22 PM

I dont know what i'm doing wrong?:

sin^-^1 \frac{2.74}{3} = 1.151

**ADY** · August 24th 2009, 10:25 PM

I dont know what i'm doing wrong?:

$sin^-1 \frac{2.74}{3} = 1.151$

**ADY** · August 25th 2009, 01:48 AM

Originally Posted by mathaddict

How did you get 1.2 by the way .

to find the inverse sine, from your calculator , press shift , sin , then you will see sin^{-1} appearing on the screen , then you key in (2.74/3) to get 65.97.

Got it! - Your working in Degree's and i'm working in Radians!

**ADY** · August 25th 2009, 08:39 AM

How do i find the lengths of DC DB and therefore find BC

**masters** · August 25th 2009, 10:10 AM

Originally Posted by ADY

How do i find the lengths of DC DB and therefore find BC

Refer to Running-gag's drawing.

$(AC)^2=(AD)^2+(AD)^2$

$DC=\sqrt{(AC)^2-(AD)^2}$

${\color{red}DC=\sqrt{8^2-(8\sin 20)^2}}$

$(AB)^2=(AD)^2+(DB)^2$

$DB=\sqrt{(AB)^2-(AD)^2}$

${\color{red}DB=\sqrt{3^2-(8\sin 20)^2}}$

${\color{red}BC=DC-DB}$

**ADY** · August 25th 2009, 10:32 AM

Masters you are truely excellent at this! Bit off topic have you been doing it long?

Going back to finding the angle of ABD, thats 66 ° so is ABC = 180° - 66° = 114° ?

Thank you !

**masters** · August 25th 2009, 10:45 AM

Originally Posted by ADY

Masters you are truely excellent at this! Bit off topic have you been doing it long?

Going back to finding the angle of ABD, thats 66 ° so is ABC = 180° - 66° = 114° ?

Thank you !

Yes....rounded off.

**ADY** · August 25th 2009, 11:23 AM

Do you make the area = $8.2cm^2$

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