Guide :

# Angles 2 and three are vertical. Angle 1 = 4x-26. Angle 2 = 3x+4

Find the measure of the the angles and show all work please

## Research, Knowledge and Information :

### angles vertical angles - That Quiz

Vertical angles are formed when 2 lines intersect. ... add up to 180 0. add up to 90 0. Angles 1 & 2 are: 1. 4. 2. 3. vertical. adjacent & supplementary. ... m⊾2=3x ...

### Finding Angles - CEEMRR.COM

Finding Angles. Intersecting Lines. 1. Consecutive angles are supplementary: Angles 1 and 2 add to 180 o. 2. Vertical angles are congruent: Angles 1 and 3 are congruent.

### 2 = 6x 1 and the m 4 = 4x 17 then find the m 3 4 If m 1 = 9x ...

... 1 and the m 4 = 4x + 17, then find the m 3 4) ... 3.6 Congruent Angles Theorem 3-1 Vertical Angle Theorem Vertical ... 3 and the m 3 = 3x - 14, then find the m 3 ...

### MATH 11008: Angles

MATH 11008: Angles Angle: ... \1 and \3 are vertical angles; \2 and \4 are vertical angles. ... 2. (x+1) (4x - 56) 3. (3x - 15) (5x + 11) 4. (5x + 3) (4x + 6) 5.

### Angles page #1 Angles - lsrhs.net

26. Given:! ∠ETR is a straight angle.!9.!27. ... m∠1=3x+y+5!! m∠2=10y−4x−2!! m∠3=2x+2y−2 Find:! 2 ... Vertical Angles are equal.! 1 3 2 4

### 3 - Angles - Blackboard

... applying the Angle Bisector Theorem, 3x 4x = DC 24, so DC= 18. ... all angles (call them \1, \2, \3, and \4) ... Since a is vertical to the alternate interior ...

## Suggested Questions And Answer :

### how to solve this using graphical method..

It's not clear what is to be solved. Nevertheless, the three points a, b and c can be plotted on Cartesian coordinates based on a vertical axis running North to South (top to bottom or positive to negative) and a horizontal axis running West to East (left to right or negative to positive). The distances given I understand to be distances from the origin and the angles represent the angles between lines joining the origin to each of the points and one or other of the axes. East of South is the angle between the negative y axis and the line on the positive side of x; N30W is the angle between the line and the positive y axis on the negative side of x; S of W is the angle between the line and the negative x axis on the negative y side. Point a becomes (5cos30, -5cos60); point b is (-7cos60, 7cos30); c is (-10cos70, -10cos20). All dimensions are cm because N=1cm. The trigonometric ratios are: cos60=0.5, cos30=sqrt(3)/2, cos70=0.342 and cos20=0.940. Sqrt(3)=1.732. All decimals to three places accuracy. Joining the three points together we get a triangle. The three sides of the triangle represent three lines which could be represented by linear equations involving x and y. So the three points of intersection are a, b and c which evaluate to (4.330,-2.5), (-3.5,6.062), (-3.420,-9.397).

### how do find the size of sides in an octogon, when nothing is given?

how do find the size of sides in an octogon, when nothing is given? the size of the piece is 2.250x2.250. There are eight vertices in an octagon, so there are eight internal angles from the center to the vertices. Each angle is three hundred sixty degrees divided by eight, or forty-five degrees. We can use the tangent of half that angle to find the length of one half of a side of the octagon (as shown in the figure). The base of the angle is one half of the square that encloses the octagon. x / 1.125 = tan 22.5 x / 1.125 = 0.414213 x = 0.414213 * 1.125 x = 0.46599 The length of the side is twice that. s = 2 * x s = 2 * 0.46599 s = 0.93198

### Geometry problem - explain all steps taken to solve

The base AC, which is a chord of the circumscribed circle, can be bisected and the perpendicular bisector will meet the apex B, because triangle ABC is isosceles and the bisector splits the triangle into two congruent right-angled triangles. The bisector of chord AC also passes through the centre of the circle, O. BO=AO=OC=r, radius of the circle. Angle AOC is twice ABC because the angle at the centre is twice the angle at the circumference for sector ABC. AB=BC=12m. Angle ABC is bisected by the side bisector on AC. If AC's bisector is at X on AC, then AX=XC=10m and sinABX=AX/AB=10/12=5/6. Angle AOX=twice angle ABX [or AOX=180-twice angle ABX], because angle AOC=2ABC and ABC=2ABX and AOC=2AOX. It follows that angle AOX=ABC=2sin^-1(5/6). AX/AO=10/r=sinAOX.  Trig identities: sin2y=2sinycosy and sin(y)=sin(180-y). Applying the first of these identities: sinAOX=sinABC=sin2ABX=2sinABXcosABX=2*5/6*sqrt(1-(5/6)^2). Therefore r=10/(2sinABXcosABX)10/(2*5/6*sqrt(1-(5/6)^2))=10.8544m approx. (36/sqrt(11) or 36sqrt(11)/11). (In actual fact AOC as the angle at the centre is a reflex angle and ABC is an obtuse angle. This does not alter the logic because the sine of an angle is the same as the sine of its complement. BO, the radius, is longer than BX. Angle AOX=180-2ABX, so AX/AO=sin(180-2ABX)=sin(2ABX).)

### GEOMETRY HELP WILL MARK AS BEST ANSWER

The blanks are ...6...angles...sides. An equiangular hexagon has congruent angles. An equilateral hexagon has congruent sides. There is no difference between the three. c) true; d) false e) false f) true. No diagram, so we don't know what the labels represent. Angle B is a right angle, because the slope of AB is (1-4)/3-(-1))=-3/4 and slope of BC is (-3-1)/-3=4/3. When two slopes multiplied together = -1 the angle between them is 90 degrees. Alternatively, Pythagoras' theorem applies: AB^2=(1-4)^2+(3-(-1))^2=25; BC^2=(-3-(-1))^2+3^2=25; AC^2=(-3-4)^2+1=50; therefore AC^2=AB^2+BC^2 and AB=BC=5, so it's an isosceles right-angled triangle.

### What is the value of <CAD if <ABC=120 degrees?

If a regular hexagon is inscribed within a circle of diameter of 12cm (radius 6cm) then, since a hexagon is effectively made up of 6 equilateral triangles, the sides of the hexagon are equal to one another, and are equal to the radius of the circle. If the sides of the hexagon are length 5cm, the hexagon is inscribed within another circle of diameter 10cm. So the picture I'm getting is of two concentric circles, centre O, the inner of which circumscribes the hexagon. The locations of points A, B, C and D are not specified in the question. The angles of a regular hexagon are fixed regardless of its size, each interior angle being 120 degrees. The angle ABC=120 degrees suggests that A, B and C are three consecutive vertices of the hexagon. The point D I see as being located on the larger circle as an extension of OB. The triangle ABC is isosceles with the equal sides being two 5cm sides of the hexagon. The height of ABC is half the radius of the inner circle, so its length is 2.5cm. The difference between the radii (OD-OB) of the two circles is 1cm, so the length of line DX (X is the mid-point of AC) from D to AC is 2.5+1=3.5cm. We know that angle BAC is 30 degrees. AX=sqrt(AB^2-BX^2)=sqrt(25-6.25)=4.33cm (Pythagoras). The tangent of angle CAD=DX/AX=3.5/4.33=0.81. Angle CAD=tan^-1(0.81)=38.95 degrees.

### Find the interior angle of the triangle with vertices A=(3,2) B=(4,5) C=(-1,-1)

There are three interior angles, all of which can be found. First, the length of the sides. CA=5, because we can use Pythagoras and the coordinates and of points A and C: AC^2=(2-(-1))^2+(3-(-1))^2=9+16, so AC=sqrt(25)=5. Similarly, BC=sqrt(61) and AB=sqrt(10), using the coords of B and C, and A and B. Using the cosine rule, we can find angle A: a^2=b^2+c^2-2bccosA: 61=25+10-2*5sqrt(10)cosA; cosA=-26/(10sqrt(10))=-0.8222, A=145.30 degrees. We can now use the sine rule to find angle B: sinB/b=sinA/a: sinB=5*0.5692/sqrt(61)=0.3644, B=21.37 degrees. Angle C=180-(A+B)=13.33 degrees.

### Angles 2 and three are vertical. Angle 1 = 4x-26. Angle 2 = 3x+4

????????????? y=3x+4 ??????? that dont be vertikal...slope=3...thats bout 71.565 deg

### cast rule behavior

360 degrees is divided into 4 quadrants: Q1: 0-90; Q2=90-180: Q3=180-270; Q4=270-360. In Q1, all trig functions are positive; in Q2, only sine is positive (sin(X)=sin(180-X)); in Q3, only tangent is positive and in Q4 only cosine is positive. CAST=cosine,all,sine,tangent=Q4,Q1,Q2,Q3. Graphically cosine and sine look similar but they are displaced by a phase difference of 90 degrees. Tangent resembles neither cosine nor sine, but is nevertheless periodic in that the pattern repeats. The red curve is y=sin(x), blue is y=cos)x), green is y=tan(x) (asymptotes are shown as green vertical lines). Between the y-axis and the first green line to the right all functions are positive (Q1), where the red and green curves intersect further to the right, we have Q2, Q3 is where the green curve is positive, up to the next vertical green line, and Q4 goes off the picture, but Q4 is also up to the first green vertical line to the left of the y-axis, followed further left by Q3, etc. The regular pattern continues indefinitely repeating every 360 degrees. The gap between the green verticals is 180 degrees.

### how do you get three kite shapes into an equilateral triangle

First draw an equilateral triangle. From one of the vertices, A, drop a perpendicular, AX, to bisect the opposite side BC at X. From X draw a line to meet AB at P, the midpoint of AB. Draw a line PQ parallel to BC to meet AC at Q. This line will intersect AX at right angles at point Y. AYX should be in a straight line. Join X to P and X to Q. Draw a line RS parallel to PQ and BC so that R on side BA is about 1/3 the way along BP from B, and S is correspondingly 1/3 the way along CQ from C. RS meets AYX at Z, so AYZX are on the same line. From P and Q drop perpendiculars on to BC at M and N respectively. V is the point where PX intersects RS, and W is the point where QX intersects RS. RVWS is a straight line.  The quadrilaterals AQZP, PVMR and QSNW are kites. The quadrilateral AQXP is an equilateral rhombus joining the three kites with common sides PV (on PX) and QW (on QX), and AP and AQ of AQZP.

### if three parallel lines are cut by transversal how many pairs of vertical angles are formed

2*3=6 ................