 Hi and welcome to the session. Let's work out the following question. The question says A and B be two fixed points in a horizontal line at a distance 50 cm apart. Two fine strings AC and BC of length 30 cm and 40 cm respectively support a weight W at C. Show that the tension in the string CA and CBR in the ratio 4 is to 3. So let A and B be two points in a horizontal line at a distance 50 cm. So this is 50 cm. Two fine strings AC and BC of length 30 cm and 40 cm respectively they're supporting the weight W at the point C. We have to show that the tensions T1 and T2 in the string CA and CBR in the ratio 4 is to 3. Let's start with the solution to the question. Now here we see that this angle is 90 degree or we can say this is a right angle triangle because 30 square plus 40 square is equal to 2500 that is same as AB square. It means that ABC is a right angle triangle at the point C. Now let angle ABC be theta that we have taken this angle to be theta. Now this angle is theta therefore angle BAC is 90 degree minus theta. Now sine theta is equal to AC divided by BC sorry this is AC divided by AB. This is equal to 90 by 50 that is equal to 9 by 5. Sorry this is 30 divided by 50 that is equal to 3 by 5. And we see that sine 90 degree minus theta is equal to BC upon AB that is equal to 40 upon 50 and that is equal to 4 by 5. Or we can say that cos theta is 4 by 5. Now applying Lamy's theorem point C we get T1 divided by sine 90 degree plus theta is equal to T2 upon sine 180 degree minus theta is equal to W upon sine 90 degree. This happens because the angle between T2 and W that is this and this will be 180 degree minus 90 plus theta that is 90 degree plus theta. Angle between T1 and W that is this and this is 180 degree minus theta and angle between T1 and T2 is 90 degree. This we have already found so this is how we get this. This implies T1 upon cos theta is equal to T2 upon sine theta and this is equal to W upon 1. This implies T1 is W cos theta and T2 is W sine theta. This implies T1 is W into now cos theta we found out was 4 by 5 and T2 will be W into sine theta sine theta was 3 by 5. Now dividing this by this we have T1 upon T2 will be equal to W into 4 upon 5 divided by W into 3 upon 5. W gets cancelled, 5 gets cancelled we have 4 upon 3. So T1 is to T2 is 4 is to 3. So this is what we were supposed to prove in this question. I hope that you understood the solution and enjoyed the session. Have a good day.