I'm in college and I knew this before, but I'm having a severe brain fart and I can't think straight. It's an understandable result of studying statics.
>pic related
If I were to split any right angled triangle into two parts from point C, how do I know the distance between point C and AB, and the distance between the point where C meets AB to A and B?
I know this is basic high school shit, but yeah, I forgot.
>>7701993
use law of cosines to solve for angle C and then use law of sines to solve for angle A, and from there use ambiguous case of law of sines
Make a Pythagorean triangle or do the law of sins.
sinA/ a = sinB / b = sinC / c
It's not too hard, anon.
a^2+b^2=c^2. Draw a line from C to the center of c making two 90° angles. Now you have a^2=1/2b^2+the new line^2
Use tangent of theta
>>7701993
www.wikihow.com/Find-the-Height-of-a-Triangle
>>7701997
Funny enough our trig class just covered this, so this for sure.
>>7701997
Or he could use the law of cosines to solve for theta straight away, then use sine of theta to get the solution.
>>7701993
its a right anged triangle, right? By the law of sines, the right angle is at C, so the sides 30cm and 40cm are attached to this right angle. Using the fact that a rectangle with width 30cm and length 40cm is 1200 cm^2 in area, and the fact that splitting such a rectangle across its diagonal produces 2 congruent right triangles (because rectangles have 2 pairs of parallel sides) with side lengths 30cm, 40cm, (applying the pythagorean theorem) 50cm, we can say that 1200cm^ 2 is twice the area of your triangle, so it is 600cm^2. Doing something similar, if we cut the triangle ABC into 2 parts with an altitude, and add 2 congruent parts with their respective positions, we can make a rectangle with 50cm width, ? cm length and 1200cm^2 area. With simple division, we get 24cm, the altitude between C and AB is 24cm.
>>7701993
you can compute the area of this triangle in at least two ways.
Either it's AC*BC/2, which is A = 30*40/2 = 600cm^2
OR, if you call H the orthogonal projection of C on (AB), then the area is also A = CH*AB/2 = 25*CH
So you have an equation with CH as an unknown, which is the distance you are lookign for.
25CH = 600 -> CH = 24.
Yes, you do know that since the triangle is at least isosceles. If you added a line from to the segment below and did so to make two 90deg angles, you would make a perfect bisection of the segment. Any non-symmetric appropriations would require substantiation on your part. Meaning you would have to bring a plausible clause. Like in medium difficulty sudoku.
I'm sorry you came to 4chan for help on this.
>>7703615
Oh wait, oops I didnt even open your pic. I thought it showed they were both 50 in length. Good thing I brought some substantiation with me. Nevertheless, you could always posit a point of bisection, you just can't do much more with it without first intergrating that info in to this particular triangle.
Angle C is obviously 90 degrees, just work your way from there.
