tu5l12-right-triangle-trigonometry-given-a-ratio-algebra-2-precalulus

# Interactive video lesson plan for: TU5L12 Right triangle trigonometry given a ratio algebra 2 precalulus

#### Activity overview:

Good day students on this clip we are going to go over how to use a given ratio to find another trig ratio okay so write this down using a trig function to find other trig functions alright alright so let us take a look at an example where we are going to use one trig function to find the value of another trig function number one if sine theta equals negative three over five find the value of the other five trig functions. We are going to assume theta is in standard position in quadrant three so what we are going to do in this case is we are going to use this trig function to generate a right triangle and then we are going to use SOHCAHTOA to finish off if we want to generate a right triangle sine theta is equals to negative three over five what ration is this? we know from sohcahtoa that sine is opposite over hypotenuse so if we look at this the top is the opposite over the bottom which is the hypotenuse so remember that the hypotenuse is always the positive so we are going to create a right triangle in quadrant three and we are going to use this as the opposite and that as the hypotenuse so what we are going to do is draw the coordinate system this is my x and this is my y and the opposite let me draw my right triangle first i am going to draw the opposite something like this make this the opposite and that is going to be the hypotenuse right there okay so the opposite is negative three so this is going to be negative three going down negative three units I am sorry put negative three right here and this entire length right here is going to be negative three going down negative three and the hypotenuse is going to be 5 well how do I know this because the angle is always starting from the x axis going down so there goes my theta and if this is theta that automatically makes this side my opposite right so this is the longest side which is my hypotenuse alright and then this is my adjacent alright so in order to find all six trig ratios we need to find what my adjacent is alright so i can use the I need to find this side adjacent right here so I am going to go I am going to use the pythagorean theorem a^2+b^2=c^2 so i can call this opposite I can call this a and then call this b and then call this right here c alright so lets plug it in and then we are going to find what b is which is the adjacent so a^2 is going to be -3 squared plus b squared equals 5 squared so we have nine plus b squared equals twenty five subtract nine from both sides and then you have b equals twenty five minus nine is sixteen b squared plus sixteen equals that b is four alright so but since we are going back in this case b is going to be negative negative four because it is going back the left side so b is going to be negative four okay alright so lets find out what the other trig ratios are I already know sine theta is negative three over five and from CAH cosine is adjacent over hypotenuse so adjacent is negative four divided by the hypotenuse which is five and uhhm tangent is opposite over adjacent which is negative three over negative four which divides we have minus over minus you end up with three fourths so thats the final answer now for the reciprocal trig function sine goes with cosecant the reciprocal of sine is cosecant so cosecant thetat is negative five over three you reciprocate cosine you get secant secant theta is negative five over four and then tangent theta is cotangent theta if you reciprocate tangent you get cotangent which is four over three and there goes your final result okay basically you are just using the ratio to create a triangle and then uhm from there you use sohcah toa to finish off the problem. alright so thanks for paying attention to this clip and be sure to subscribe to mathgotserved.com

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