RiTri 01
Simplify radicals, including rationalizing denominator. A radical is in its simplest form when 1. no perfect square factor other than 1 is under the radical sign, 2. no fraction is under the radical sign, and 3. no fraction has a radical in the denominator. To simplify radicals, factor out the number under the square root symbol in a perfect square and another number. Take the root of the perfect square and put it outside the square root symbol. For example, square root(60)=square root(4x15)=2xsquare root(15). To factor radicals out of the denominator multiply the fraction by the radical (from the denominator) over the radical. For example, 4/square root of 5 = 4x(square root 5)/(square root 5)x(square root 5)= 4x(square root 5)/4. RiTri 02 State Pythagorean Theorem. The Pythagorean Theorem states that in a right triangle with legs a and b and hypotenuse c, a squared plus b squared equals c squared. RiTri 03 Use Pythagorean Theorem to solve problems / find missing sides. To use the Pythagorean theorem, simply set it up (a^2+b^2=c^2) and plug in the known side lengths, then solve algebraically for the missing side lengths. For example, if one leg is 3 and the hypotenuse is 5, then 9+b=25, so x=16. RiTri 04 Fill in missing side of common Pythagorean Triplets. Two common Pythagorean triplets are 3-4-5 and 5-12-13. Therefore, 6-8-10, 9-12-15, 30-40-50, etc and 10-24-26, 50-120-130, etc are also triplets. So, it you see any two of these numbers, you can fill in the others. RiTri 04h Know and apply Plato's formula to find Pythagorean Triplets. Plato's formula is (taking n), 2n, n^2-1, and n^2+1 for each side of the triangle. This formula is just a simplified version of Euclid's (on the left). RiTri 05 Prove Pythagorean Theorem using area. First, use four identical right triangles to create a square within a square, with the inside square made up of all side c (see the diagram on the left). The area of the bigger square is a^2+2ab+b^2. The area of the c square plus the triangles is c^2+2ab. These are equal to each other, so a^2+2ab+b^2=c^2+2ab. Simplify and the Pythagorean theorem is found: a^2+b^2=c^2. RiTri 06 Determine whether a triangle is right, obtuse, or acute. If c is the longest side length, and a^2+b^2=c^2, then the triangle is right. If c is the longest side length and a^2+b^2>c^2, then the triangle is acute. If c is the longest side length and a^2+b^2<c^2, then the triangle is obtuse. RiTri 07 State the ratios of sides for special right triangles (30-60-90, 45-45-90). For 30-60-90 triangles, the leg opposite the angle measured 30 degrees is a. The leg opposite the angle with a measure of 60 degrees is a(square root of 3). The hypotenuse is 2a. For 45-45-90 triangles, the legs are both equal to a and the hypotenuse is equal to a(square root of 2). RiTri 07h Derive the ratios of sides for special right triangles. RiTri 08
Use special right triangles to solve problems. To use 30-60-90 and 45-45-90 triangle relationships to solve problems, plug in the values you know, and solve algebraically for the unknown sides. In some cases, it is necessary to change the form of a side length to make it equal to _(root3) or _(root2). RiTri 09 Find the area of regular polygons, given apothem and side. The apothem of a polygon is the altitude from the center point to the side. To find the area of a regular polygon, use the formula, A=1/2(side)(apothem)(number of sides). A=1/2(perimeter)(apothem) is equivalent and can also be used. RiTri 10 Find the area of a regular polygon, given apothem or side (triangle, quadrilateral, or hexagon). To find the apothem/side, use special right triangles. Make the triangle by using half the side and the apothem as the legs and connect the center to the vertex (radius) for the hypotenuse. In triangles, the central angle of the center is 60 degrees, in hexagons it is 30 degrees and in quadrilaterals it is 45 degrees. In hexagons, the side will always be equal to the radius. Use RiTri 07 and 08. Once you have both the apothem and side length use A=1/2(a)(s)(n) to find the area of the regular polygon. RiTri 11 Derive the distance formula from scratch. Begin with the Pythagorean theorem. If a right triangle triangle with legs a and b and hypotenuse c, then a^2 + b^2 = c^2. Take the diagram on the left. Since it forms a right triangle, we can substitute the corresponding values into the Pythagorean theorem. The legs are x2-x1 and y2-y1 and the hypotenuse is d, or distance. So, (x2-x1)^2 + (y2-y1)^2 = d^2. Use the symmetric property to say that d^2=(x2-x1)^2 + (y2-y1)^2. Then, square root both sides to get d=square root of [(x2-x1)^2 + (y2-y1)^2]. This is the distance formula. RiTri 12 Use the distance formula, simplify result, to find distance. The distance formula is d = square root of [(x2-x1)^2 + (y2-y1)^2]. To use it, plug in the coordinates you know. The first coordinate is (x1,y1) and the second is (x2,y2). Then simplify using your algebra skills to find what d (distance) is equal to. RiTri 13 Write the equation for a circle, given center and radius (and vice-versa). The equation for a circle is (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius. As (x,y) is a point on the circle, you can use those values in conjunction with either the center or radius to find the other, either radius or center. RiTri 13h Drive the equation for a circle, given center and radius. Begin with the Pythagorean theorem. If a right triangle triangle with legs a and b and hypotenuse c, then a^2 + b^2 = c^2. Take the diagram on the left. (h,k) is the center of the circle and (x,y) is a point on the circle. r is the radius. Since a right triangle is formed, we can substitute the corresponding values into the Pythagorean theorem. The legs are (x-h) and (y-k) and the hypotenuse is the distance between the center and a point on the circle, otherwise know as the radius. After substitution, (x-h)^2 + (y-k)^2 = r^2. This is the equation for a circle. RiTri 99 Solve problems you haven't seen before, using analysis and synthesis of the information learned so far. I think I can. Can you? |
Trig 01
Identify trig ratios by sides (SOH, CAH, TOA). Use the diagram on the left. SOH stands for sine: opposite side divided by hypotenuse. CAH stands for cosine: adjacent side divided by hypotenuse. TOA stands for tan: opposite side divided by adjacent side. Trig 02 Given three sides of right triangle, find TanA, SinA, and CosA as ratios. Use the diagram to the left. The sine ratio (SinA) is the length of the opposite side divided by the hypotenuse (a/h). The cosine ratio (CosA) is the length of the adjacent side divided by the hypotenuse (b/h). The tan ratio (TanH) is the length of the opposite side divided by the adjacent side (a/b). Trig 03 Given a special right triangle, find TanA, SinA, CosA (30-60-90, 45-45-90) by hand (no calculator, exact, simplified radical answers). Using RiTri 07 and 08, assign values to the sides. Then use SOH, CAH, TOA and simplify to find the sine, cosine, and tangent for each angle. For example, if the sides of a 30-60-90 triangle are 5, 5(roots), and 10, then Tan(30°)=5/5(root3)=root3/3, Sin(30°)=5/10=1/2, and Cos(30°)=5(root3)/10=root3/2. Tan(60°)=5(root3)/5=root3, Sin(60°)=5(root3)/10, and Cos(60°)=5/10=1/2. If the sides of a 45-45-90 triangle are 5, 5, and 5(root2), then Tan(45°)=5/5=1, Sin(45°)=5/5(root2)=root2/2, and Cos(45°)=5/5(root2)=root2/2. The Sin, Tan, and Cos ratios will always be the same for the angle measures. Trig 04 Given an angle and side in a right triangle, find missing sides using algebra and calculator. First, from the perspective of the known angle, identify which sides are given (opposite and/or hypotenuse and/or adjacent). Next, identify which trigonometry function (sine, cosine, or tangent) uses the sides given (use SOH, CAH, TOA). Third, write out the equation (ex: Sin(x°)=opposite/hypotenuse)then solve algebraically. Depending on what the directions say, use a calculator to solve and round to whatever digit it asks for. Trig 05 Find the area of a regular pentagon, octagon, dodecagon, using Trig and a calculator. Answers should be accurate/precise to the nearest 1/100th. Using trig, all that is needed to find the area of a regular polygon is the number of sides (n) and either the apothem or side/half the side. You will use RiTri 10 and A=1/2 x apothem (a) x side (s) x number of sides (n). First, find half the central angle (figure 1) by calculating 360/2n or 180/n. The tan of this angle measure is equal to the opposite side (half the side length) divided by by the adjacent side (the apothem). Plug in the angle measure you found and the known length, then use algebra to solve. Either save this exact value in your calculator or plug in the fraction for a or s (double for s) in A=(1/2)asn and solve. If you are given the side length (s) and the number of sides (n), you can use (s^2 x n)/(4 x tan(180/n)) to find area (derived in figure 2). If you are given the apothem (a) and the number of sides (n), you can use a^2 x n x tan(180/n) to find area (derived in figure 3). To solve, plug in the values and use order of operations and your calculator to find the answer. Trig 06
Solve problems to find height using angle of elevation/depression using trig. First, let's define angle of elevation and depression. The angle of elevation is the angle above the horizontal (like looking up at a flag vs straight ahead. The angle of depression is the angle below the horizontal (like looking at a boat from the top of cliff vs straight ahead). To solve problems (often word problems), create a right triangle using the height you are trying to find as one of the legs and the horizontal as the other. Then use Trig 04 to find the missing side/height. An example is on the left. Trig 07 Given two sides, find missing angles using algebra and calculator. First, set up the proportion (sin of x=opposite/hypotenuse, cos of x=adjacent/hypotenuse, or tan of x=opposite/adjacent). Since both sides are known, the variable will be next to the trigonometry function. To solve for the variable, you must reverse the trig function. To do so, multiply both sides by the inverse of the trig function (often the 2nd of the same button on a calculator). This will isolate the variable, then use the calculator to find the missing angle, rounding to the appropriate decimal. Trig 08 Use trig identities to simplify expressions. There are three categories of trig identities: reciprocal, Pythagorean, and quotient. The first reciprocal identity states that 1/sine of an angle = cosecant of that angle (csc). The second reciprocal identity states that 1/cosine of an angle = secant of that angle (sec). The third reciprocal identity states that 1/tangent of an angle = cotangent of that angle (cot). The first Pythagorean identity states that sine squared of an angle + cosine squared of that angle = 1. The second states that 1 + cotangent squared of an angle = cosecant squared of that angle. The third states that tangent squared of an angle + 1 = secant squared of that angle. The first quotient identity states that the tangent of an angle = sine of that angle / cosine of that angle. The second states that the cotangent of an angle = cosine of that angle / sine of that angle. To simplify expressions, use algebra and substitution of these identities. Occasionally, writing all functions in terms of sines and cosines, or combing fractions by getting a common denominator, or breaking one fraction into two, or factoring may help. Trig 09 Know and apply Law of Sines. The Law of Sines says that the sine of one angle in a triangle over the length of the opposite side is equal to the sine of another angle in the triangle over the length of the opposite side is equal to the sine of the last angle over the length of the opposite side. Given the triangle in the figure "Law of Sines" to the left, SinA/a = SinB/b = SinC/c. Given two angles and one side, you can find the remaining angle and sides. Use the triangle sum theorem to find the third angle. Then take two of the proportion at a time, plug in the known values and solve for the unknown using algebra. Trig 10 Derive the Law of Sines. Begin with triangleABC, with sides a, b, and c (side is opposite corresponding angle, shown in figure "Law of Sines"). Draw the altitude from side b to angle B, calling it h. Since two right triangles are created, we can say SinA=h/c and SinC=h/a, simplifying to h=c x SinA and h=a x SinC. According to the transitive property, we can set these two values together, saying c x SinA = a x SinC. Divide both sides by ca and SinA/a = SinC/c. Now return to the original triangle and draw the altitude from side a to angle A, calling it j. Using the same process as above, SinB/b = SinC/c. Again using the transitive property, we can say SinA/a = SinB/b = SinC/c, which is the Law of Sines. Trig 11 Know and apply the Law of Cosines. The Law of Cosines states that in triangleABC with sides a, b, and c (side is opposite corresponding angle, shown in figure "Law of Sines" and in Trig 12), a^2 = b^2 + c^2-2bcCosA and b^2=a^2+c^2-2acCosB and c^2=a^2+b^2-2abCosC. It can be used when you are given three sides (SSS) or two sides with an angle in between (SAS). Apply it by plugging in the known values and solving algebraically for the unknown variables. Trig 12 Derive the Law of Cosines. The Law of Cosines has three versions. This example derives one version and the others can be derived by drawing the altitude from a different angle, then substituting the corresponding sides. Use the image below on the left to help with the proof/derivation. Trig 13
Solve triangles. Be able to determine whether 0, 1, 2, or infinitely many triangles can be formed with the given three pieces of information (angles and/or sides). To solve triangles, determine whether to use law of sines (LoS) or cosines (LoC), then plug in the known values and solve algebraically for the variables. If you are given three angles (AAA), you cannot use either LoS or LoC because infinitely many triangles can be made. If you are given three sides (SSS) or two sides with an angle in between (SAS), use LoC and exactly one triangle can be formed. If you are given two angles with a side in-between (ASA) or two angles and a following side (AAS), use LoS and exactly one triangle can be formed. If you are given one angle and two following sides (ASS), use LoS, but it is ambiguous so 0, 1, or 2 triangles can be formed. To determine how many triangles can be made, solve for both the acute (first) and obtuse (second) triangles. If both work, then 2 triangles can be made. If one works, then 1 triangle. If neither work, then 0 triangles. Trig 99 Solve problems you haven't seen before, using analysis and synthesis of the information learned so far. I think I can. Can you? |
CIRC 01
Identify radius, diameter, chord, secant, tangent, circumference, central angle, arc, intercepted arc, minor arc, major arc, measure of arc, semicircle, length of arc. Define circle, and name it correctly by center point. The terms above are labeled in the image to the left. A circle is the set of points in a plane that are a given equal distance from a given point in the plane. A circle is named with a single letter, the letter of the center point. Example: if the center point is point B, then the circle made of the set of points a given distance from point B is named Circle B. CIRC 02 Given central angle, find the measure of the arc intercepted. The measure of the intercepted angle of a central angle is the same as the measure of the central angle. CIRC 03 Given central angle and radius, find length of arc intercepted. The length of an arc is found by dividing the measure of the arc or central angle by 360 then multiplying that by the circle's circumference (or, 2pi(radius)). CIRC 04 Use algebra to solve for missing radius or central angle, given arc length and other. Use the formula from CIRC 03 ((arc measure/360)x2pi(radius)), then plug in the known values and use algebra to solve for the unknown values. CIRC 05 Define inscribed angle and given an inscribed angle, find the measure of arc intercepted. An inscribed angle is the angle created by two chords (or parts of secants) that share an endpoint/vertex on the circle. The measure of the arc intercepted is twice the measure of the inscribed. CIRC 06 Apply Right Angle and Arc Intercept corollaries to solve problems. The Right Angle corollary states that if the endpoints of an inscribed angle lay on the endpoints of a diameter, then the inscribed angle is right, or the measure of the inscribed angle is 90 degrees. The Arc Intercept corollary states that is two inscribed angles share an intercepted arc, then the inscribed angles are congruent. The Inscribed Quadrilateral corollary states that if a quadrilateral is inscribed in a circle, then the opposite angles are supplementary. CIRC 07 Classify circle problems with three criteria: measures or lengths, secants or tangents, where is vertex?
CIRC 08
Complete a summary of angle arc relationships table like page 596. If the angle is formed by two radii with the vertex in the center of the circle, then the measure of the angle is equal to the measure of the arc. If the angle is formed by a tangent and secant with the vertex on the circle or two chords/parts of secants with the vertex on the circle, then the measure of the angle is equal to half the measure of the arc. If the angle is formed by two intersecting chords/parts of secants with the vertex inside the circle, then the measure of the angle is equal to the average of the measures of the intersected arc and the intersected arc of the vertical angle. If the angle is formed by a secant and a tangent, or two secants, or two tangents with the vertex outside the circle, then the measure of the angle is equal to half the opposite arc measure minus the closer arc. CIRC 09 Apply Chords and Arcs theorem to solve problems. This theorem says that congruent central angles have congruent chords, congruent chords have congruent arcs, and congruent arcs have congruent central angles. To apply this theorem, first identify what you need to find out and what information you have. Then select the applicable form of this theorem and plug in your values to solve. CIRC 10 Solve problems to find measures of angles and arcs involving secant and tangent, intersection ON the circle. The measure of the angle formed by the secant and tangent is equal to half the extreme inscribed angle. Formula is in CIRC 08. CIRC 11 Solve problems to find measures of angles and arcs involving two secants (or chords) intersecting IN the circle's interior. The middle angles equal the average of the intersected angle and the intersected angle of the vertical angle. Plug in and solve. Formula is in CIRC 08. CIRC 12 Solve problems to find measures of angles and arcs involving two secants/tangents intersecting OUTSIDE the circle. The measure of the angle formed by the two secants/tangents intersecting outside the circle is equal to the measure of the inscribed arc (farther arc from the angle) minus the closer arc. Plug in and solve. Formula is in CIRC 08. CIRC 13 Prove that the segments of two intersecting tangents to a circle are congruent. First, draw two radii that intersect the tangents. A radii will always intersect a tangent at a right angle (CIRC 18) and two radii of the same circle are always congruent. A segment can be drawn connecting the center of the circle with the point of intersection between the tangents and because of the reflexive property of congruence, that segment is congruent to itself. The two triangles formed are congruent by RHL and the segments of the tangents are congruent by CPCTC. CIRC 14 Solve problems to find the lengths of segments formed by tangents (vertex outside). The lengths of the segments formed by tangents that intersect are congruent. Plug in and solve. CIRC 15 Solve problems to find the lengths of segments formed by secants (vertex outside). The whole part of the secant times the part outside the circle equals the whole part of the second secant times the part outside the circle. Plug in and solve. CIRC 16 Solve problems to find the lengths of segments formed by tangents and secants (vertex on). Use what you know about the relationships between tangents, radii, secants, etc to help solve such problems. Use of right triangles is also helpful (such as special right triangle side relationships and the Pythagorean Theorem). CIRC 17 Solve problems to find the lengths of segments formed by intersecting chords, including chords perpendicular to diameter. The segment parts times each other equal the other segment's parts times each other (ex: with endpoints on circle a, b, c, and d and POI z, az x bz = cz x dz). Plug in and solve. CIRC 18 Solve length problems using radius intersecting tangents and chords. A radii that intersects a chord at a right angle bisects the chord and the arc. A radii will always intersect a tangent at a right angle. Use these concepts to plug in and solve. CIRC 19 Complete a summary of relations of lengths for secant, tangent, and chord segments (Like page 607). This skill is asking for a summary table of relationships, just like in CIRC 08 except with lengths instead of measures. Instead of rewriting it all, I will just refer you to CIRC 13 - 18. CIRC 99 Solve problems you haven't seen before, using analysis and synthesis of the information learned so far. I think I can. Can you? |