Geometry help? How do I find the values of x and y?
Here's a homework problem I'm stuck on: Find the values of x and y m - the measure of < - angle sign In triangle JKL, segment JK is congruent to segment KL, m
Geometry: Find the values of x and y?
FInd the values of x and y. 1)http://i90.photobucket.com/albums/k262/pinklotus_2006/geo-1.jpg 2)http://i90.photobucket.com/albums/k262/pinklotus_2006/geo1-1.jpg 3)http://i90.photobucket.com/albums/k262/pinklotus_2006/geo2-1.jpg Ok...so how would i set the equation up to solve it? thnx:)
How do i find the values of x and y in geometry?
The triangle has... 3x+y+7+90=180 solve it for me please. I did it but I dun know if its correct! I get stuck at parts where its like x+y=180 idk what to do then
Geometry Question - find value of x and y (trapezoid and triangles)?
if the ratios of two sides of two triangles are proportionate, and the angle in between the two sides is common to both the triangles, the traingles are similar. and the ratio of the third sides of the traingle is same as that of the ratio of the other sides of the triangles. hence, (x+3)/36=1/3 => x=9 substituting in (x+y)/36= 2/3 => y=15. Is that right?
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
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
How do you find the value of x in a scalene triangle?
That depends on what I know at the outset, but the Law of Cosines and the Law of Sines are great tools for finding unknown sides and angles in scalene triangles. Dropping an altitude from a vertex to its opposite side gives two right triangles to ponder, too. Other useful tools are Heron’s formula for the area of a triangle given the three side lengths, as well as the fact that the product of two side lengths and the sine of the angle between them gives twice the area of the triangle.
If [math]x^2+y^2=16 \; \text{&} \;[/math][math] [/math][math]xy=9[/math]. What is the value of [math]x+y[/math]?
This can be done byx^2+y^2=16or (x+y)^2–2xy=16or (x+y)^2=16+2xyor (x+y)^2=16+2.9or (x+y)^2=34:. (x+y)=(+/-)√342.Also other than this formula two more formulas can be applied.i. since, x^2+y^2=16or 2(x^2+y^2)=16.2or (x+y)^2+(x-y)^2=32…..(1)ii. also, xy=9or 4xy=9.4or (x+y)^2-(x-y)^2=36…..(2)By eliminating (2) from (1) we get,[{(x+y)^2-(x-y)^2=36} + {(x+y)^2+(x-y)^2=32}]or 2(x+y)^2=68or (x+y)^2=34or (x+y)=(+/-)√34ONE COMMENT HERE WHICH IS NOTABLE IS THAT THE NUMBER OF VALUES OF (x+y) WILL ALWAYS BE 2. THIS IS BECUASE OF THE FACT THAT IS WE ARE DEALING WITH BINOMIAL EQUATION THAT IS X RAISED TO THE POWER OF 2( IN SIMPLE) SO THE NUMBER OF VALUES OF X AND Y WILL ALWAYS BE MORE THAN ONE SO THEIR SUM WILL ALSO HAVE MORE THAN ONE VALUES.