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Students in this video, let's start analyzing two important triangles, this will help us retain the

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important sine cos andante values already over here have Triangle ABC and the angles of triangle ABC

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are forty five degrees, 45 degrees and 90 degrees.

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Now in Triangle Park you are the anchors are 60 degrees, thirty degrees and 90 degrees.

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So we have marked the angles over here.

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Now let's see Triangle ABC.

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You can see that these two angles are equal to we haven't isosceles right triangle over here.

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So therefore if this is of X units, this side also will be of X units and hence AC will be of root

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of X squared plus X squared units.

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Right.

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That's equal route root 2x.

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Therefore the ratio of sites of abs to is to AC will be of the form.

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One is two, one is to root two.

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All right.

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But we need to remember this, this will help us a lot.

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Now let's see triangle P.

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Q are the ratio of sides of triangle P.

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Q Are will be in the form root three over here, one over here and two over here.

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All right.

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Now what are the things that we can note the side opposite the bigger angle will be bigger.

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Right over here you have root three and over here you have one through Rotary's opposite sixty degrees.

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So it's the bigger value and one is opposite thirty degrees.

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Now the value of Rotary's around one point seven.

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Right.

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So it's greater than one hundred and ninety degrees.

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The biggest angle.

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So you have two opposite ninety degrees.

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All right.

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Now we can see that this is a right angle triangle.

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Right.

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Therefore over here, if you square this, you have one square and if you square root three, you have

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three.

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Right.

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The one plus three gives you four.

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And when you take root of four.

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Right, you get that equal to two.

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And that's the two over here.

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So it's following the Pythagoras Theorem property.

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All right.

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Now, if we remember these two triangles, that is if you remember the ratio of the sides in the case

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of these two triangles, we can easily remember a lot of the important sine cos and tan ratios.

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All right.

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So let's keep this aside.

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OK, now let's get on with the important sign costs and then values, and you need to remember this,

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and this is a memory aid for you to remember these values.

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Now, let's start filling this table.

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What would be signed 30.

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Now you have a 30 degree angle over here, right?

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The sign is opposite by hypotenuse.

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So that would be one divided by two.

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Right.

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So sign 30 is equal to one by two.

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And we write that over here.

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What about cost 30?

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Now, again, we have the 30 degree angle over here causes a descent by hypotenuse.

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That's Route three, divided by two.

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Right.

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Similarly, what is done down is opposite by Addison.

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That would be one by route.

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So let's write these values.

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We have cost 30 as Route three by two and then 30 as one by Route three.

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All right.

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Now let's get to 45 five.

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What is St. Peter?

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Its opposite by hypotenuse.

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So if this is forty five, sine forty five is one by route.

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Right, let's right that over here.

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What would be cost.

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Forty five.

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It's a by hypotenuse that's also equal to one by two.

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And what is ten.

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Forty five.

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It's opposite by Addison.

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That's one by one.

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And you get it as one.

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So let's write these values over here.

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Cost forty five is one by two and then forty five is equal to one.

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All right.

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Now what about sixty degrees.

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What is sine sixty.

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You have a 60 degree angle over here.

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Sine is opposite by adjacent, opposite by hypotenuse.

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So that would be route three by two.

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Right.

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We are Route three by two over here.

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Similarly 60 will be one by two.

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Right.

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That's adjacent by hypotenuse.

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We have one by two over here.

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And what would be ten.

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Sixty, that's route three divided by one.

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That's equal to Route three right now.

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Let's go to the zero degree angle.

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Now, imagine you have a triangle over here and the sides are X, Y, and this side is each.

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And this angle over here is data.

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Now, you can imagine that when Tita becomes zero, right, this value of X becomes zero.

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Right.

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As you keep reducing T to the value of X will keep producing.

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And when two is equal to zero, X will be equal to zero.

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Right now what is s.A.

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That's X by X, right?

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90 days equal to in that triangle, 70 days X by H.

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Now if X is equal to zero, this becomes zero.

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That's why you signed zero is equal to zero.

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All right.

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Now what about cost zero.

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Let's think it in this manner.

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In this case, when detail is zero, right.

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It becomes equal to white.

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That is why it becomes equal to X.

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These two values are equal.

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Right.

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And what is costing over here?

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Because that is why you buy it.

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And because these are equal, you get cost.

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It is equal to one.

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Right.

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All right.

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Now let's proceed to Dante de de de de de de divided by Costeja.

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That's zero divided by one, and that is equal to zero.

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Right?

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So we have got the values of sine cos and tan data when data is equal to zero.

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All right, let's proceed.

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Now what about the case?

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Twenty two is equal to ninety degree.

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When Peter is equal to ninety degrees, this X will become equal to its right.

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You can think of it in this manner.

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As you keep increasing data rate, the value of X will keep increasing and it is approaching the value

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of X, right.

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So when data is equal to 90 degree X will be equal to X.

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Right.

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Therefore S.E De which is X by x rayed, that would be one because these two values are equal.

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So sine 90 is equal to one.

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What about cost ninety at this case, where to is ninety, you can see that Y will be equal to zero.

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Right.

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You can think of it in this manner as this keeps increasing, this keeps decreasing.

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But when X is equal to its right is just one straight line over here.

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That is why is equal to zero now cost is equal to Y by each so that would be zero by eight which is

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equal to zero to cost.

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Ninety is equal to zero and and data is equal to sign because that's one divided by zero.

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Right.

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And that is equal to infinity or it is not defined because division by zero is not defined.

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Let me write that again.

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Division by zero is not defined now or here.

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We have got the important properties and with this you will be able to solve any trigonometry question.