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Q: In triangle 2 angle x equal?
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How do you calculate sine?

The sine of an angle of a right triangle - which is a triangle containing one 90o angle - is calculated as the length of the side opposite the angle divided by the length of the hypotenuse. For very small values of x, sin(x) is approximately equal to x.


The vertex angle of an isosceles triangle is 9 degree more than a base angle find the angles of the triangle in mathimatics?

An isosceles triangle has 2 angles that are equal. All angles to a triangle = 180 degrees so you set it up as a algebra equation as X+X+X+9=180 3X=171 in this X = 57 So 2 angles are 57 and one is 57+9= 66 Sorry about the spelling.


How do i find the value x in a scalene angle?

If it is a scalene triangle then: 180 - the sum of the 2 known angles = x angle


An isocleles triangle has one angle witch measures 66 what is the measures of the other two?

In an isosceles triangle there are two equal sides and two equal angles. In a triangle ABC, if angle A is between the sides of equal length, then angles B and C are equal.Without knowing which angle (A, B or C above) is 66o there are two possible answers:If the 66o angle is between the two sides of equal length (angle A) then the other two angles (B and C) are (180o - 66o) / 2 = 57o each.If the 66o angle is not between the sides of equal length (angle B or C), then the other two angles are 66o (the other angle of C and B) and (angle A) 180o - 66o x 2 = 48o.


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It is 1/2 of length x width


How do you find the area of an isosceles right triangle with hypotenuse 6sqrt2?

As it's an isosceles right triangle, the right angle is the angle between the two sides of equal length. Using Pythagoras, the lengths of these sides, and hence the area can be found: 2 x side2 = (6√2)2 ⇒ side2 = 62 ⇒ side = 6 ⇒ area = 1/2 x 6 x 6 = 18 sq units.


How do you find the side length of a triangle?

The most popular triangle used in construction, engineering and mathematics is the right triangle, which has a 90o angle at the base. We'll call your base "x", height "y" and diagonal (hypotenuse) "r".You can find these values using your calculator's trigonometric functions, or you can use the Pythagorean Theorem.Finding Values (Pythagorean Theorem):The Pythagorean Theorem is straightforward and states that, for a right triangle, r^2 = x^2 + y^2. We can use a little bit of algebra to find x, y and r.r = sqrt(x^2 + y^2)r^2 = x^2 + y^2r = sqrt(x^2 + y^2); remove the ^2 from rx = sqrt(r^2 - y^2)r^2 = x^2 + y^2r^2 - y^2 = x^2; move y^2 to the left side of the equationsqrt(r^2 - y^2) = x; remove the ^2 from xy = sqrt(r^2 - x^2)r^2 = x^2 + y^2r^2 - x^2 = y^2; move x^2 to the left side of the equationsqrt(r^2 - x^2) = y; remove the ^2 from yFinding Values (Trigonometric Functions):Acquiring Ratios:cos(angle) = x/rsin(angle) = y/rtan(angle) = y/xAcquiring Angles:cos-1(x/r) = anglesin-1(y/r) = angletan-1(y/x) = angleThat looks confusing. What are cos, sin, tan, etc?They're functions. You give a function input and it outputs something else. How they do this is complicated and required calculus.x = cos(angle) * rcos(angle) = x/rcos(angle) * r = x; move r to the left side of the equationy = sin(angle) * rsin(angle) = y/rsin(angle) * r = y; move r to the left side of the equationx = 1 / (tan(angle)/y)tan(angle) = y/xtan(angle) / y = 1/x; move y to the left side of the equation1 / (tan(angle) / y) = x; flip the equationy = tan(angle) * xtan(angle) = y/xtan(angle) * x = y; move x to the left side of the equationr = 1 / (cos(angle)/x)cos(angle) = x/rcos(angle) / x = 1/r; move x to the left side of the equation1 / (cos(angle)/x) = r; flip the equationr = 1 / (sin(angle)/y)sin(angle) = y/rsin(angle) / y = 1/r; move y to the left side of the equation1 / (sin(angle)/y) = r; flip the equation


Calculate the largest angle of a triangle in which one angle is four times each of the others?

120 (& 2 x 30)


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Slope = Rise/Run = y/x 1. leg = rise = y 2. leg = run = x 3. hypotenuse = √(x^2 + y^2) tan (angle 1) = x/y angle 1 = arctan(x/y) tan (angle 2) = y/x angle 2 = arctan(y/x)


If half of an angle is equal to the twice of its supplementary angle Then find the measure of angle-?

1/2 x = 2 (180-x) x = 4(180-x) x = 720 - 4x 5x = 720 x = 144 degrees


Show that sin x equals cos x for some angle t between 0 and 90?

sin(x) = cos(x)sin(x)/cos(x) = tan(x) = 1x = arctan(1) = 45 degreessin(45)=cos(45) = Sqrt(2)/2 Answer: By observation. Since Sine = Opposite over hypotenuse and Cosine = Adjacent over hypotenuse. Any right angle triangle where the opposite and adjacent sides are the same length will have Sine equal to Cosine. This only happens with an isosceles triangle (two sides are equal in length). When one angle is 90o the other two are 45o.