Algebal review 6 pps

Triangle Trigonometry There are special ratios we can use when working with right triangles. They are based on the trigonometric func- tions called sine, cosine, and tangent. For an angle, ⌰, within a right triangle, we can use these formulas: sin ⌰ϭ ᎏ hy o p p o p t o e s n i u te se ᎏ cos ⌰ϭ ᎏ hy a p d o ja t c e e n n u t se ᎏ tan ⌰ϭ ᎏ o ad p j p a o c s e i n te t ᎏ The popular mnemonic to use to remember these formulas is SOH CAH TOA. SOH stands for Sin: Opposite/Hypotenuse CAH stands for Cos: Adjacent/Hypotenuse TOA stands for Tan: Opposite/Adjacent Although trigonometry is tested on the SAT, all SAT trigonometry questions can also be solved using geom- etry (such as rules of 45-45-90 and 30-60-90 triangles), so knowledge of trigonometry is not essential. But if you don’t bother learning trigonometry, be sure you understand triangle geometry completely. opposite hypotenuse adjacent hypotenuse opposi te adjacent To find sin ⌰ To find cos ⌰ To find tan ⌰ ⌰ ⌰ ⌰ –GEOMETRY REVIEW– 119 TRIG VALUES OF SOME COMMON ANGLES SIN COS TAN 30° ᎏ 1 2 ᎏ 45° 1 60° ᎏ 1 2 ᎏ ͙3 ෆ ͙3 ෆ ᎏ 2 ͙2 ෆ ᎏ 2 ͙2 ෆ ᎏ 2 ͙3 ෆ ᎏ 3 ͙3 ෆ ᎏ 2 Example First, let’s solve using trigonometry: We know that cos 45° ϭ , so we can write an equation: ᎏ hy a p d o ja t c e e n n u t se ᎏ ϭ ᎏ 1 x 0 ᎏ ϭ Find cross products. 2 ϫ 10 ϭ x͙2 ෆ Simplify. 20 ϭ x͙2 ෆ ϭ x Now, multiply by (which equals 1), to remove the ͙2 ෆ from the denominator. ϫϭx ϭ x 10͙2 ෆ ϭ x Now let’s solve using rules of 45-45-90 triangles, which is a lot simpler: The length of the hypotenuse ϭ ͙2 ෆ ϫ the length of a leg of the triangle. Therefore, because the leg is 10, the hypotenuse is ͙2 ෆ ϫ 10 ϭ 10͙2 ෆ . 20͙2 ෆ ᎏ 2 20 ᎏ ͙2 ෆ ͙2 ෆ ᎏ ͙2 ෆ ͙2 ෆ ᎏ ͙2 ෆ 20 ᎏ ͙2 ෆ 20 ᎏ ͙2 ෆ ͙2 ෆ ᎏ 2 ͙2 ෆ ᎏ 2 ͙2 ෆ ᎏ 2 45° x 10 –GEOMETRY REVIEW– 120  Circles A circle is a closed figure in which each point of the circle is the same distance from the center of the circle. Angles and Arcs of a Circle ■ An arc is a curved section of a circle. ■ A minor arc is an arc less than or equal to 180°. A major arc is an arc greater than or equal to 180°. ■ A central angle of a circle is an angle with its vertex at the center and sides that are radii. Arcs have the same degree measure as the central angle whose sides meet the circle at the two ends of the arc. Central Angle Major Arc Minor Arc –GEOMETRY REVIEW– 121 Length of an Arc To find the length of an arc, multiply the circumference of the circle, 2πr,where r ϭ the radius of the circle, by the fraction ᎏ 36 x 0 ᎏ , with x being the degree measure of the central angle: 2πr ϫ ᎏ 36 x 0 ᎏ ϭ ᎏ 2 3 π 6 r 0 x ᎏ ϭ ᎏ 1 π 8 rx 0 ᎏ Example Find the length of the arc if x ϭ 90 and r ϭ 56. L ϭ ᎏ 1 π 8 rx 0 ᎏ L ϭ ᎏ π(5 1 6 8 ) 0 (90) ᎏ L ϭ ᎏ π( 2 56) ᎏ L ϭ 28π The length of the arc is 28π. Practice Question If x ϭ 32 and r ϭ 18, what is the length of the arc shown in the figure above? a. ᎏ 16 5 π ᎏ b. ᎏ 32 5 π ᎏ c. 36π d. ᎏ 28 5 8π ᎏ e. 576π x° r r r x° –GEOMETRY REVIEW– 122 Answer a. To find the length of an arc, use the formula ᎏ 1 π 8 rx 0 ᎏ ,where r ϭ the radius of the circle and x ϭ the meas- ure of the central angle of the arc. In this case, r ϭ 18 and x ϭ 32. ᎏ 1 π 8 rx 0 ᎏ ϭ ᎏ π(1 1 8 8 ) 0 (32) ᎏ ϭ ᎏ π 1 (3 0 2) ᎏ ϭ ᎏ π ( 5 16) ᎏ ϭ ᎏ 16 5 π ᎏ Area of a Sector A sector of a circle is a slice of a circle formed by two radii and an arc. To find the area of a sector, multiply the area of a circle, πr 2 , by the fraction ᎏ 36 x 0 ᎏ , with x being the degree meas- ure of the central angle: ᎏ π 3 r 6 2 0 x ᎏ . Example Given x ϭ 120 and r ϭ 9, find the area of the sector: A ϭ ᎏ π 3 r 6 2 0 x ᎏ A ϭ ᎏ π(9 2 3 ) 6 ( 0 120) ᎏ A ϭ ᎏ π( 3 9 2 ) ᎏ A ϭ ᎏ 81 3 π ᎏ A ϭ 27π The area of the sector is 27π. x° r r secto r –GEOMETRY REVIEW– 123 Practice Question What is the area of the sector shown above? a. ᎏ 4 3 9 6 π 0 ᎏ b. ᎏ 7 3 π ᎏ c. ᎏ 49 3 π ᎏ d. 280π e. 5,880π Answer c. To find the area of a sector, use the formula ᎏ π 3 r 6 2 0 x ᎏ ,where r ϭ the radius of the circle and x ϭ the measure of the central angle of the arc. In this case, r ϭ 7 and x ϭ 120. ᎏ π 3 r 6 2 0 x ᎏ ϭ ᎏ π(7 2 3 ) 6 ( 0 120) ᎏ ϭ ᎏ π(49 3 ) 6 ( 0 120) ᎏ ϭ ᎏ π( 3 49) ᎏ ϭ ᎏ 49 3 π ᎏ Tangents A tangent is a line that intersects a circle at one point only. tangent point of intersection 120° 7 –GEOMETRY REVIEW– 124 . angle: 2πr ϫ ᎏ 36 x 0 ᎏ ϭ ᎏ 2 3 π 6 r 0 x ᎏ ϭ ᎏ 1 π 8 rx 0 ᎏ Example Find the length of the arc if x ϭ 90 and r ϭ 56. L ϭ ᎏ 1 π 8 rx 0 ᎏ L ϭ ᎏ π(5 1 6 8 ) 0 (90) ᎏ L ϭ ᎏ π( 2 56) ᎏ L ϭ 28π The. what is the length of the arc shown in the figure above? a. ᎏ 16 5 π ᎏ b. ᎏ 32 5 π ᎏ c. 36 d. ᎏ 28 5 8π ᎏ e. 5 76 x° r r r x° –GEOMETRY REVIEW 122 Answer a. To find the length of an arc, use the. ᎏ 36 x 0 ᎏ , with x being the degree meas- ure of the central angle: ᎏ π 3 r 6 2 0 x ᎏ . Example Given x ϭ 120 and r ϭ 9, find the area of the sector: A ϭ ᎏ π 3 r 6 2 0 x ᎏ A ϭ ᎏ π(9 2 3 ) 6 ( 0 120) ᎏ A