1. ## Triangles!

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

2. Hi

If I have well understood the situation is the following

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

3. Well understood!

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

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

5. 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?
$\displaystyle \sin \angle ABD = \frac{2.74}{3}$, so find $\displaystyle \arcsin \frac{2.74}{3}$ to find the measure of $\displaystyle \angle ABD$.

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

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

7. Originally Posted by ADY
so how do inverse sine? $\displaystyle sin^-1 \frac{2.74}{3} = 1.2$ ?
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.

8. I dont know what i'm doing wrong?:

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

9. I dont know what i'm doing wrong?:

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

10. 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!

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

12. Originally Posted by ADY
How do i find the lengths of DC DB and therefore find BC
Refer to Running-gag's drawing.

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

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

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

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

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

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

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

13. 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 !

14. 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.

15. Do you make the area = $\displaystyle 8.2cm^2$

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