1. Right Triangles and Trigonometry By Mathew Chacko and Jessica Wu illuminati

2. Overview of Concepts ● Sine, Cosine, and Tangent help provide ratios that find the interior angles and side lengths of triangles. ● SOHCAHTOA use when right triangle ● first, you must use an angle that is not right (90) ● determine whether to use sin, cos, or tan based on the side lengths you know (opp and hyp=sin, adj and hyp=cos, adj and opp=tan) ● put sin,cos, or tan in front of said angle measure and set equal to trig ratio

3. ● find a proportion using unknown angle and trig ratios (sin, cos, tan) ● tan(A)=opp/adj, sin(A)=opp/hyp, cos(A)=adj/hyp ● if using sine, find the opp and hyp of said angle, same for cosine and tangent ● after finding a fraction using given numbers, apply inverse to the sign ● example below (easy) Inverse SOHCAHTOA A 16 20 find measure of <A use tan b/c you know opp and adj side compared to <A tan(A)=16/20=4/5 apply inverse tan-1(4/5) is about 38.7 <A≈38.7 degrees

4. Law of Sine ● A,B,C are angle measurements ● a,b,c are side lengths ● law of sine is sinA/a= sinB/b= sinC/c ● first proportion you use must have given a side and angle ● proportion you are solving for must have only one known side or angle ● angle measure over side set equal to angle/side (unknown) ● then cross multiply to isolate unknown variable then solve ● remember it only finds acute angles

5. Law of Cosine ● law of cosine is a²=b²+c²-2bc cos(A), b²=a²+c²-2ac cos (B), c²=a²+b²- 2ab cos (C) ● one side squared set equal to other two sides squared minus two times those sides multiplied together than cosine of angle corresponding with original side ● only need to use if only know all the sides or SAS ● example below (medium) A B C 60° 10 a 8 FIND THE MISSING SIDE a²=64+100-2(80)cos(60) 164-80=84 a²=2√21 ≈9.2 a=9.2 now that you know all the sides and an angle you can solve the the rest of the angles using law of sines

6. Two Triangles, One, or None ● must test if a SSA (butt) triangle ● example, if you use sine and find an angle to be 30 degrees, you must also test the theory using 180 minus 30 which equals 150 b/c sine only finds acute angles ● by using side lengths, determine if one neither or both are plausible ● if a>b then A>B, if both answers work, then two triangles, if only one than one triangle ● no solution if sine ends up greater than one ● example below (difficult) c a=10 b=12 A=24° B C FIND THE OTHER TWO ANGLES SSA triangle so solve for more than one possibility sin24/10=sinB/12=sinB= sin24/10 <B≈29 degrees or 151 degrees based on triangle, 12>10 so it could be either so two triangles 180-24-29=127, 180-151-24=6 final angle= 6 or 127 degrees A=24 degrees B=29 or 151 degrees C=6 or 127 degrees

7. Connections to Other Units ● trigonometry relates to proportions b/c students make proportions with sine, cosine and tangent ● example would be sin 62/32=sin76/b sin62(b)=sin76(32) b=33.5 ● also people use SOHCAHTOA when you don’t know two sides of a right triangle ● if two sides are given, then just apply pythagorean theorem ● which connects to unit 7a ● you may sometimes need to use trigonometry to find the area/volume of a triangle in a 3-D or 2-D shape

8. Common Mistakes ● many students forget the ratios for sine, cosine, and tangent a simple and easy trick to help memorize these confusing ratios would be the word “SOHCAHTOA” pic on the right ● students also forget to solve for more than one solution when you have a “butt” triangle or side side angle triangle (SSA) ● remember in previous units when we prove triangles congruent we never use SSA so you must solve for all possible solutions if SSA

9. Real Life Scenario You are building a ramp for a moving company that slants upward at a certain angle degree. You know that the side on the ground is 10m and the slant upwards is 12m. You also know that the angle on the ground is 24°. Find the rest of the dimensions. A B C c a=12m b=10m 24° Work: (use law of sine) 1. sin24/12= sinB/10 (multiply the proportionality) 2.sin24x10= sinBx12 (use calculator to reduce equation) 3.angle B is about 19.8 degreees (not 180-19.8 b/c a>b) 4.180 - two <’s= 136.2 degrees 5.sin 24/12=sin 136.2/c (cross multiply) 6.c is about 20.4 a=12 b=10 c=20.4 <A=24 <B=19.8 <C= 136.2

10. GOOD LUCK ON YOUR FINAL