/sci/ - Science & Math

bump, this is excruciatingly hard.

Work out the edges for the main triangle, then use similar trianges to work out the little triangle. This gives AD, then take AD from AB

use pythagoras to calc AC.

Then calc the angle in B.

then calc using the angle and BC AD.

Damn this shit aint hard.

You use trigonometry.

still nobody answered my question about which side is the height.
if i wanted to find the area would it be: 1/2(AB*CD) or 1/2(AB*AC) ?

>>2851582
The "height" of a triangle is not well-defined. You need to use the law of cosines and the pythagorean theorem.

>>2851570

You're so bad. You can't use pythag for that. You don't have any sides for ABC.


Use law of cosines

AB= 14
BC= 12

and Angle C is 90.

Use law of cosines to find angle A. After that add angle A+D and subtract that from 180 to find the third angle in triangle ADC. Subtract that missing angle from 90 and you now have the missing angle from DBC.

Use law of sines or cosines. Done.

>>2851597
You can use the pythagorean theorem. |AC|^2=|AB|^2-|BC|^2. Then you can find cosB, whence |BD|=|BC|cosB.

Is ADC similar to ACB or DCB?

>>2851610

I guess that works if they are similar.

>>2851624
If what is similar? I'm not using any similarity. |BD| is the projection of |BC| onto |AB|.

Anyway OP, I went ahead and did it. The answer is B. Use pythagoras and then the sine rule a couple of times

Start by finding the length of AC. You do this using the pythagorean theorem to get sqrt(52).

Now you can use the law of sines to find angle B:
14 = sqrt(52) / sin(angle B)
sin(angle B) = sqrt(52) / 14
Using inverse sine, angle B is approximately 31 degrees.

Using law of sines again, we can find length CD:
CD / sin(31) = 12
Thus, CD is approximately 6.18

Now we have the lengths of one leg and the hypotenuse of a right triangle. Use pythagorean theorem to find the 2nd leg AD and you get approximately 3.72 .BD = AB - AD

14 - 3.72 = 10.28. This is closest to answer B. YAY MATHS

to OP: both of CD and AC are heights:
CD is a height relatively to AB
AC is a height relatively to CB

>>2851658

but which side would i use to find the area of triangle ACB?

>>2851677

CD. To visualize this, place a rectangle around the triangle and you'll see that this length corresponds with the height of the rectangle.

Ok thanks for all the responses

BC is the geometric mean of BD and AB.

that is to say:

(BD/BC)=(BC/AB)

which can be rearranged thusly:

BD = (BC^2)/AB

BD = (12^2)/14

BD = 144/14

BD = 72/7, (B)

>>2851736

(samefag here)

for extra points, now that you have BD you can use the same geometric mean technique to find CD:

(AD/CD) = (CD/BD)

CD^2 = AD*BD

CD = (AD*BD)^(1/2)

CD = ((52/14)*(144/14))^1/2

CD = 6.18094504

>>2851756

with AB and CD (which you can assign as your base and height, respectively), the area is:

Area = (1/2)b*h
Area = (1/2)AB*CD
Area = (1/2)*(14)*(6.18094504)
Area = 43.2666153 square units

Hope this helped. I know that previous anons used law of sine/cosine, but the geometric mean of an altitude dropped from the right angle of a triangle is something taught (and usually forgotten) at least 2 years before those other laws, and simplicity is best for me!

>>2851736

>mfw OP uses law of sines on his 9th grade geometry HW and his teacher knows he cheated

>>2851526

dot producttttttttt