SSLC MATHS:IMPORTANT PROBLEMS FOR THE UPCOMING EXAMS-PART 03

Added: 2026-05-08

[00:00:08]Hello everyone. Today we are going to solve some of the important problems which uh which will be helpful for the upcoming exams.

[00:00:18]So the first problem is in the figure find the value of seeant B and C A.

[00:00:28]So first of all we should write the given values which we need to find. So first one is seeant B.

[00:00:37]We need to find seeant B. So definition of seeant A B is adjacent divided by hypotenuse. So this is the definition of seeant B. Adjacent by hypotenuse. Now we have to consider the angle from which we have to find out the value. So the angle here we have to consider is angle B. So while considering the angle B we need to find the different sides of the given triangle. So if you consider angle B then opposite to this angle B will be this opposite side. So this will be the opposite side adjacent to 90°. This will be the adjacent side and the remaining is hypotenuse.

[00:01:25]So seeant B is equal to so here from the figure adjacent side is five divided by hypotenuse is 13. So seeant B from the given figure is 5 by3.

[00:01:40]Now we need to find C A.

[00:01:44]So the general definition of C A is adjacent side divided by opposite side. So now we have to find the values through the given angle. So now we have to consider the angle A. So through this angle we need to find the different sides. So if you consider angle A then opposite to this angle A is this will be the opposite side considering the angle A this will be the adjacent side and the hypotenuse side will remain as it is.

[00:02:22]Now C is equal to so from the given side we find that the adjacent side is 12 divided by opposite side is five. So 12 x 5 is the answer for a.

[00:02:48]Now we are going for the second problem.

[00:02:51]In the figure if the measure of the hypotenuse is twice the opposite side of the angle alpha then find the value of cos alpha. Now we have to first consider the figure and consider the angle alpha and find the different sides of the given triangle.

[00:03:10]So if you consider this angle alpha then opposite to this angle alpha this will be the opposite side. Now adjacent to 90° this will be the this will be the adjacent side and the remaining will be the hypotenuse side.

[00:03:30]So from the given question it is clear that side a c is equal to 2 * bc.

[00:03:44]So they have given that hypotenuse is twice the opposite side. So hypotenuse is a c twice the opposite side is bc.

[00:03:54]Now we can take any of the trigonometric ratios and solve the given problem. Let us take the sign ratio. We have sin alpha. Sin alpha is equal to opposite by hypotenuse. This is the general definition.

[00:04:15]Sin alpha is equal to So opposite side is bc divided by hypotenuse side is twice dc.

[00:04:28]So sin alpha equ= bc by 2 bc. So bc gets cancelled. Now you have sin alpha is equal to 1 by 2.

[00:04:40]Now comparing with the given side we get opposite by hypotenuse.

[00:04:46]Now we can use the pythogoras theorem.

[00:04:50]Using Pythogoras theorem we have opposite square plus adjacent square is equal to hypotenuse square.

[00:05:11]So now we have opposite side is 1 square plus adjacent side. We need to find out hypotenuse is 2² adjacent square is = 2² is 4. Shift 1 square to the right side + 1 will become -1.

[00:05:30]Adjacent square is = 4 - 1 is 3.

[00:05:34]Now you have adjacent is equal to taking the square root. So adjacent is equal to square root of 3.

[00:05:45]So we need to find the cos alpha. So by definition cos alpha is equal to adjacent by hypotenuse.

[00:05:56]So cos alpha is equal to adjacent is <unk>3 divided by hypotenuse. We have 2.

[00:06:05]So cos alpha is equal to <unk>3x2.

[00:06:09]So this is the final answer. We have cos alpha is equal to <unk>3 by 2.

[00:06:37]Now we are going for the third problem.

[00:06:40]In the figure, angle ABC is equal to 90°. Write the values of the following.

[00:06:46]First one is sin alpha. Second one is tan theta. So first we need to find sin alpha.

[00:06:56]So first we need to take the general definition of sin alpha. So sin alpha is equal to opposite by hypotenuse.

[00:07:08]So now we have to consider the angle alpha and find the given sides of the triangle. So considering the angle alpha opposite to this alpha this will be the opposite side.

[00:07:22]Considering the angle alpha adjacent to the 90° this will be the adjacent side and the remaining side will be hypotenuse.

[00:07:31]So sin alpha is equal to opposite side is 3 divided by hypotenus is 5. So sin alpha is equal to 3x 5.

[00:07:45]Now we need to find the second value that is tan theta.

[00:07:51]First we need to write the definition of tan theta that is equal to opposite by adjacent side.

[00:08:00]Now we have to consider the angle.

[00:08:03]So here the angle considered is angle theta with respect to theta. We need to name the sides. So this is angle theta opposite to angle theta. This will become opposite side adjacent to 90° with respect to theta.

[00:08:18]This will be the adjacent side.

[00:08:22]Hypotenuse will remain as it is.

[00:08:25]So we have tan theta is equal to opposite side is 4 divided by adjacent side is 3. So tan theta is equal to 4x3.

[00:08:56]Now we are going to solve the fourth problem. Find the HCF of 135 and 75 by frame factorization method and then find the LCM of 20 and HCF of 135 and 75.

[00:09:13]So first let us find the HCF of 135 comma 75.

[00:09:23]So first of all we need to find the prime factors of 135 that is equal to we have 3 into 3 into 3 3's are 9 9 3's are 27 27 into 5 27 into 5 we get 135 that is equal to write in the powers of the given prime factors. So you have 3 to the^ of 3 into 5 ^ of 1.

[00:10:08]So this is the these are the prime factors of 135.

[00:10:14]Now we need to find the prime factors of 75 that is equal to 3 into 5 into 5 5 are 25 25 3's are 75. Now write the factors with respect to the power 3 ^ of 1 into 5 ^ of 2.

[00:10:50]Now to find the HCF Now to find the HCF of 135, 75, we need to get the factors with respect to the lower power. So the common power factors prime factors here is three in both the sides three. So with the lowest power so that will be only 3 ^ of 1 into now the common factors common prime factors in both the numbers you have five we need to write the factor with the lowest power. So that is 5 to the^ of 1.

[00:11:36]So HCF of 135A 75 is equal to 3 3 5 are 15.

[00:11:47]So HCF of 135A 75 is equal to 15.

[00:11:54]We need to find the LCM of 20 and HCF of 135, 75.

[00:12:23]So first the prime factors of 20 that is equal to we have 2 2's are 4 4 5 are 20. Now we need to write with respect to the power 2 ^ of 2 into 5 ^ of 1.

[00:12:48]Now we need to find the prime factors of HCF of 135A 75. So here only we have find out HCF of 135A 75 is 15. Therefore we need to find the prime factors of 15 is equal to 3 ^ of 1 into 5 ^ of 1.

[00:13:18]LCM of 20 is equal to here we need to write the powers. So here we have 2 ^ of 2 then you have 3 ^ of 1 and here you have into 5 ^ of 1. So these are the factors of 20 and 15.

[00:13:51]So 2² is 4 into 3 into 5 that is equal to 4 3's are 12 12 into 5 so that is equal to 60.

[00:14:06]Therefore the LCM of 20 is equal to 60. So which is the final answer.

[00:14:26]Very important problem. This has been asked many a times in the exam.

[00:14:42]Now going for the fifth problem. If the product of LCM and HCF of two numbers is 2366 and their HCF is 13, then find their LCM. If one of their numbers is 91, then find the other. Very important problem.

[00:15:06]We know that HCF into LCM is equal to product of two numbers that is A into B.

[00:15:34]So, HCF into LCM is equal to A into B.

[00:15:45]From the given question, we need the HCF is 13. Substituting the value of HCF 13 into LCM is equal to the product of two numbers given is 23 66.

[00:16:01]Substituting the value now cross multiply 13 we get LCM is equal to 23 666 divided by 13.

[00:16:15]Now we need to divide the given value with respect to 13.

[00:16:22]So let us divide easily 2 3 6. So here we can see 13 1 are 13.

[00:16:33]So the remainder is 10. So take one value 106.

[00:16:40]Now we can see 138 are 138 are 1 4 13 8 are 1 4 Now subtract the given value 106 - 104 is 2.

[00:16:56]Take value one down that is 26. Now 13 2's are 26. So the remainder is zero. So 2 3 6 divided by 13 we get is 182.

[00:17:10]So we have LCM is equal to 13 1 are 13 * 182. So LCM is equal to 182.

[00:17:25]So this is the first part of the answer.

[00:17:29]We need to find the LCM.

[00:17:33]Now it has given the second part also.

[00:17:36]If one of their numbers is 91 then find the other number.

[00:17:48]So we know that product of two numbers a into b is equal to hcf into lcm.

[00:18:03]So the one of the number given is 91. So 91 into B is equal to we know that HCF is 13.

[00:18:18]So 13 into LCM. Now only we find out 182.

[00:18:28]Now cross multiply the other number.

[00:18:31]B = 13 into 182 divided by 91.

[00:18:38]So clearly we can see we can divide 91 through 182 91 1 are 91 2's are 182 very easy one. So B is equal to 13 into 2. So the other number B is equal to 26.

[00:18:59]So that is the final answer.

[00:19:12]Very important problem. Kindly revise and practice and understand it.

[00:19:23]Now going for the next problem. The length of two rectangular tables are 84 cm and 10 126 cm respectively and both have a width equal to 42 cm.

[00:19:37]Find the maximum length of the square glass sheet that can exactly fit both the tables.

[00:19:45]Also calculate the total number of glass sheets required to fit each table.

[00:19:53]very tricky and important problem.

[00:19:57]So for the above question, HCF is the solution.

[00:20:02]HCF is the solution.

[00:20:10]So first of all, let the given values the length of two rectangular tables. So first rectangular table is 84 cm.

[00:20:24]Second table is 126 cm.

[00:20:30]So as we know HCF is the solution. We need to find the prime factors of the first table and second table.

[00:20:42]Prime factors of 84 cm is equal to 2 into 2 into 2 2's are 4 into 3 4 3's are 12 into so the number which we get 84 is 7 2's are 4 4 3's are 12 12 7 are 84 Now we need to write with respect to the powers. First one is 2² second one is 3 ^ 1 and third one is 7 ^ of 1. Now the prime factors of second table 126 cm that is equal to first of all we get 2 2 into 3 2 3's are 6 into again we need to write 3 2 3's are 6 6 3's are 18 into again we need to write the multiple is 7. So another multiple is 7 2 3's are 6 6 3 are 18 18 7 are 126. Now we need to write with respect to the powers 2 into 3 ^ of² into 7.

[00:22:14]So we know that htf is the solution.

[00:22:17]So therefore HCF of 84 and 126 is 2 and also we need to write the lowest power the number with the lowest common number with the lowest power. So that is 2 ^ of 1 and common factor again 3 with the lowest power 3 ^ 1 into common factor with the lowest power 7 ^ of 1. So now we need 2 3's are 6 into 7 7 6 are 42.

[00:22:56]So HCF of 84 comma 126 is equal to 42 maximum length of therefore maximum length of the square glass is equal to 42 cm.

[00:23:32]Maximum length of the square glass is equal to 42 cm.

[00:23:38]Now we need to find out the number of glass sheets required.

[00:23:45]So going for the second part you can see we need to find the glass sheets number of glass sheets required to fit each table.

[00:24:09]Number of glass sheets required to fit the first table is equal to 84 divided by 42.

[00:24:45]So here we can divide directly 42 1 are 42 2. So that is equal to two. So number of glasses required to fit the first table is two.

[00:24:58]Now, number of glass sheets required to fit the second table is equal to 26 divided by 42.

[00:25:31]So 42 1 are 42 3. So that is equal to three. So the sheets required to fit the second table is number is three. Here the number required is two. So this is the final answer.

[00:25:52]I hope you understood the all the problems today which were solved which are very important for the upcoming exams.

[00:26:01]Kindly revise it and also subscribe the channel and share the link with your friends which may be helpful for their future. Thank you.