Introduction to Trigonometry Video Lecture - (Maths) Class 10

0:0:0: have you ever wondered how ancient
0:0:2: people of Egypt were able to make
0:0:4: pyramids huge pyramids were made by
0:0:7: those people with clear and precise
0:0:10: calculations these calculations were
0:0:12: done through the concepts similar to
0:0:15: trigonometry but let us understand what
0:0:17: exactly is
0:0:19: trigonometry it is made of to Greek
0:0:21: words trigono and metry trigono means
0:0:24: triangle and metri means measure so it
0:0:28: is a study of measurement of triangle
0:0:30: angles to be more specific we study the
0:0:33: parameters of right
0:0:35: triangles the reason for studying the
0:0:37: properties of right triangles is simple
0:0:40: enough we can always divide any triangle
0:0:43: into two right
0:0:44: triangles so by studying the properties
0:0:47: of right triangles we can study all the
0:0:50: triangles so now let us first analyze a
0:0:53: right
0:0:54: triangle it has sides such as a base a
0:0:57: perpendicular and a hypotenuse it also
0:1:0: has three angles a hypotenuse is the
0:1:4: longest side of
0:1:5: triangle base is the side adjacent to
0:1:8: the angle perpendicular is a side
0:1:11: opposite to the selected
0:1:13: Angle now the first thing which comes to
0:1:16: our mind is that we can just measure
0:1:18: these values with a ruler and
0:1:20: protractor you are absolutely right if
0:1:24: we are talking about right triangles
0:1:26: which can be drawn on paper we can use
0:1:28: this method
0:1:30: however the purpose of trigonometry is
0:1:32: to solve real life
0:1:34: problems let's say we need to find the
0:1:36: height of a very tall tower will we be
0:1:40: able to measure its height the answer to
0:1:43: this is a big no but trigonometry
0:1:46: provides a solution to this problem
0:1:49: before trying to solve this let us try
0:1:51: to understand some of the basic terms
0:1:54: and functions which are used in
0:1:56: trigonometry you must have heard the
0:1:58: term ratio used in
0:2:0: mathematics we have ratios in
0:2:2: trigonometry as trigonometric
0:2:5: ratios let us study them one by one sign
0:2:9: of an angle it is the ratio of the
0:2:12: perpendicular of a triangle to its
0:2:15: hypotenuse it is represented by sin
0:2:19: Theta we should also note that sin Theta
0:2:22: is not formed through multiplication of
0:2:24: s an angle Theta it is a term that
0:2:27: represents s of an angle Theta
0:2:30: cosine of an angle it is the ratio of
0:2:33: the base of a triangle to its
0:2:36: hypotenuse it is represented by C Theta
0:2:40: tangent of an angle it is the ratio of
0:2:43: the perpendicular of a triangle to its
0:2:45: base it is represented by tan Theta
0:2:50: cosecant of an angle it is the ratio of
0:2:53: the hypotenuse of a triangle to its
0:2:56: perpendicular it is reciprocal of s s of
0:3:0: an angle it is the ratio of the
0:3:2: hypotenuse of a triangle to its base it
0:3:6: is reciprocal of cosine cotangent of an
0:3:9: angle it is the ratio of the base of a
0:3:12: triangle to its
0:3:14: perpendicular it is reciprocal of tan
0:3:17: Theta now that we have defined the
0:3:20: values we can try to find out these
0:3:23: ratios for triangle named
0:3:25: ABC here we have all these ratios as
0:3:28: shown now there is one interesting fact
0:3:31: to notice if we change the length of
0:3:34: sides of triangle there will be no
0:3:36: changes in these ratios as long as angle
0:3:39: is not
0:3:40: changed this property helps in real life
0:3:44: scenarios let us see some features of
0:3:46: these
0:3:47: ratios since the hypotenuse is the
0:3:50: longest side of the triangle the value
0:3:52: of sign and cosine are always less than
0:3:56: one similarly the value of secant and
0:3:59: cosecant is is always greater than one
0:4:2: also if any value of trigonometric ratio
0:4:4: is given we can find out the value of
0:4:7: all other
0:4:8: ratios the most important thing to do is
0:4:11: to visualize a
0:4:12: triangle let the value of sin a is 4
0:4:15: upon 5 so now first make a right
0:4:19: triangle since sin a equals
0:4:22: perpendicular upon
0:4:24: hypotenuse we take its perpendicular as
0:4:27: 4X and hypotenuse as 5
0:4:30: X I hope you guys remember the
0:4:32: Pythagoras Theorem we have hypotenuse
0:4:35: square equals perpendicular square + b
0:4:38: square now we put values in this
0:4:41: equation we get b² = 5x² - 4
0:4:47: x² base squ = 25 x² - 16
0:4:52: x² base Square = 9
0:4:56: x² so we have value of Base equals
0:5:0: 3x now we have right triangle with the
0:5:3: measurements of its sides hence we can
0:5:6: find all the required values from this
0:5:9: now you can try completing this question
0:5:12: with data that has been
0:5:14: provided let's try a bit different
0:5:16: question this time here we have a right
0:5:19: triangle ABC right angled at B here it
0:5:23: is specified that tan a is 1 find the
0:5:27: value of 2 into sin a into cos a a to
0:5:30: solve this we will first use the formula
0:5:33: tan a equals perpendicular by hypotenuse
0:5:37: for this triangle we have tan a is equal
0:5:39: to b c upon AB since tan a is 1 then BC
0:5:45: upon AB = 1 BC equal
0:5:50: AB now let us assume the ab equals BC
0:5:54: equals K now using Pythagoras Theorem we
0:5:59: get a c² = a² + B
0:6:2: c² a² = 2 k s a c = k < tk2 now we get
0:6:10: sin a = 1 upon < tk2 cos a = 1 upon <
0:6:16: tk2 now simply putting values in the
0:6:19: required
0:6:20: expression we get 2 into 1 upon < tk2
0:6:23: into 1 upon < tk2 on solving we have its
0:6:27: value equal to 1
0:6:30: now that we have solved some of the
0:6:32: questions let's try our luck with a
0:6:34: complex problem in a triangle pqr right
0:6:38: angled at q p + QR is equal to 25 cm and
0:6:43: PQ is equal to 5
0:6:45: cm determine the values of sin P cos p
0:6:49: and tan P to solve this problem let us
0:6:52: start with drawing a triangle pqr and
0:6:55: label it according to the question now
0:6:58: write down the EXT ra data given to us
0:7:1: here we have PQ = 5
0:7:4: cm label it on the
0:7:7: triangle we also have PR + QR =
0:7:11: 25 so let us first get the value of QR
0:7:15: QR = 25 minus p
0:7:19: r now since this is a right
0:7:22: triangle let us use Pythagoras Theorem
0:7:26: we get p r² = PQ ² + Q
0:7:30: r² just put the values here open the
0:7:34: brackets we get p² = 625 + p² - 50 p +
0:7:42: 25 making calculations we get 50 PR =
0:7:46: 625 +
0:7:49: 25 P = 650 upon 5050 we get p r as
0:7:55: 13 now use previous equation to find the
0:7:58: value of QR we get QR equal 12 after
0:8:4: finding these values we simply write the
0:8:6: values of trigonometric
0:8:8: ratios we get sin P = 12 upon
0:8:12: 13 cos P = 5 upon 13 tan P = 12 upon
0:8:21: 5 so now we have understood the uses of
0:8:24: trigonometric ratios to solve problems
0:8:27: related to right triangles
0:8:30: we hope that you have understood the
0:8:32: concept of
0:8:33: trigonometry now here's a secret on how
0:8:36: you can study effectively with the urv
0:8:39: app you can learn with chapter notes
0:8:41: watch video lectures and solve ncer
0:8:44: based McQ tests of this chapter on
0:8:48: urv and that's not all you also get
0:8:51: amazing courses for class 10 and much
0:8:53: more for your class 10
0:8:55: preparation thank
0:8:58: you