Find The Value Of X In A Triangle . Help Math

How can I find the values X,Y, Z of this triangle?

I have provided all the answers, and steps necessary to find the answers. However, please do not just copy the answers, but look at how I did the problems to find the answer...x=8y=16z=53.13ºRemember that if you have a right triangle and two given sides, you can apply the Pythagorean Theorem which is a^2+b^2=c^2Thus to find x, when can do 6^2+x^2 = 10^2 36+x^2 = 100100-36 = x^264=x^2x=8How can we find y now?well, we are given 30º, and we know that the other base angle is 90º, and can therefore determine that the other angle is 60º....This is a 30-60-90 triangle, and since we have one given length (x), we can determine y and the unknown triangular base length.If you remember your rules, you will know that the hypotenuse is 2x = 16 = ySo, so far we have x=8y=16Now we have to use SOCATOA to find zLets use cosine, although we could use tangent and sineCos(z) = 6/10To isolate z, you must use the inverse of cosine which can be written as Arcosine or cos^-1z = cos^-1(6/10) = 53.13º

What is the value of angle [math]x[/math] in the triangle shown in the details?

AB = AC -> 70 + a = 60 + 20 -> a = 10ABC + BCA + CAB = 180 -> 80 + 80 + b = 180 -> b = 20[BOC -> 180 - 70 - 60 = 50, thus DOE = 50, BOD = COE = 130][BDO -> 180 - 130 - 10 = 40, CEO -> 180 - 130 - 20 = 30[if you put x = OED = 60, ODE -> 180 - 60 - 50 = 70, ADE = 70 and AED = 90]SimilarlyThus AED is right angle triangle at E and is congruent with CED with x = 60

If you can see the given triangle, what will be the value of x?

I wasn't going to answer this question. But other answers are blatantly wrong, so I had to jump in.Also … I looked at this question blankly for three minutes (I should be able to solve it! My maths skills, oh God! Man, what the…) and then it finally dawned over me.Yeah, kill me.And so the answer is…..Wait, I'm not gonna tell you so easily. I take it that you are new to algebra. Well, dear lad! This is high time for you to learn something very interesting.Hence, I'll advice you too follow as I say.Take a clean sheet of paper.I mean it.Go.I am waiting.Promise, this will be worth it.Good! Thanks! Now make a huge diagram. (Like Donald Trump ’yuge!)Try to calculate all the angles that you can mark. (Easy angle sum property. Sum of all angles of a triangle is 180°. Nothing fantastic, here)Give all of them some variable name.Look at it.Realize it.Laugh at it.You'll see that there are only 4 angles that you don't know about. You gotta have the same number of variables as you have equations. If you look carefully (and if you try), there would be no more than three equations. (Comments are open, if you couldn't find the three).That means? Every freaking real number is a solution to those variables! Just put any value in [math]y[/math] and your [math]abracadabra[/math] variable will help you to find a solution for [math]x[/math]. (Hence the name) It's that simple. Yay!The only way this question had an unique solution was if at least one other angle was given to us. Others assumed the angle opposite to [math]x[/math] as [math]90°[/math], but you can't just assume things if you want to. And if any book uses the same assumption, throw it away. Because the answer for your unknown angle is:[math]x \in \R [/math][math]\text{(meaning x can be any real number)} [/math]

The sides of a triangle are given to be [math]x^2+x+1, 2x+1, x^2-1[/math]. What is the largest of the three angles of the triangle?

For the third of these side lengths to be possible (i.e. positive), we see that:[math]x^2-1>0 \implies x>1 \text{ or } x<-1[/math].For the second of these side lengths to be positive, we see that:[math]2x-1 > 0 \implies x > \frac 1 2[/math].Together, we see that [math]x>1[/math] is required. Note that when [math]x>1[/math], [math]x^2+x+1>1^2+1+1>0[/math] so the first side length will also be positive. Next, note that when [math]x>1[/math], we can multiply both sides by [math]x[/math] to see that [math]x^2>x[/math]. This implies that [math]x^2+x+1>x+x+1=2x+1[/math], and we see that the first length is longer than the second. Furthermore, when [math]x>1[/math] we also see that [math]x^2+x+1-(x^2-1)=x+2>0[/math]. So the first length is greater than the third. Finally, [math]x^2-1+2x+1-(x^2+x+1)=x-1>0[/math], so the triangle inequality is satisfied and the three sides do form a triangle.Now that we know which side is the longest, we are left to find the angle opposite this side which will be the largest angle (see David Joyce's answer which references Euclid Book I Proposition 18). We expect this angle to be written as a function of [math]x[/math]. This calculation is most easily made using the Law of cosines.[math](x^2+x+1)^2 = (2x+1)^2 + (x^2-1)^2 -2 (2x+1)(x^2-1)\cos \phi[/math]Some algebra shows that all the [math]x[/math] terms cancel out giving:[math] \cos \phi = -\frac 1 2 [/math]Since the angles of a triangle must be between zero and one hundred eighty degrees, the only valid solution to this equation is: 120 degrees = [math]\frac 2 3 \pi[/math] radiansHere's a fast way to do the algebra:First, move a term from the Right Hand Side (RHS) to the Left Hand Side (LHS):[math](x^2+x+1)^2 - (x^2-1)^2= (2x+1)^2 -2 (2x+1)(x^2-1)\cos \phi[/math]Factor the LHS using the difference of squares:[math](2+x)(2x^2+x)= (2x+1)^2 -2 (2x+1)(x^2-1)\cos \phi[/math]Factor out the [math]x[/math] on the LHS:[math]x(2+x)(2x+1)= (2x+1)^2 -2 (2x+1)(x^2-1)\cos \phi[/math]Divide by the common factor of [math]2x+1[/math] (which we can do since [math]2x+1>0[/math] is required to be a triangle).[math]x(2+x)= 2x+1 -2(x^2-1)\cos \phi[/math]Combine the terms that don't have a [math]\phi[/math]:[math]x^2-1=-2(x^2-1)\cos \phi[/math]Divide by the common factor of [math]x^2-1[/math] which is possible since [math]x^2-1>0[/math] is required for this to be a triangle:[math]1=-2\cos \phi \implies \cos \phi =- \frac 1 2[/math]