PROJECTION OF STRAIGHT LINE IN ENGINEERING DRAWING IN HINDI (SOLVED PROBLEM 4)

[@venkatrr8050]

Felt very happy about your explanation, clear and very good explanation. Even though am a teacher , like your explanation

[@sameeksha1028]

Bhut AXaa pdate ho sir ap sb axe se smj aa rha h bht bht dhnywad apka 🙏🙏

[@Sayo_03]

The wages they pay you won't fill any banks, But your hard work really matters, And you deserve a big THANKS! Thank you so much!🤗

[@_inspireverse___]

Apka channel sabse aage jayega Engineering drawing me❤

[@TIKLESACADEMY]

Thanks for your GOLDEN WORDS.... It's only possible only because of all the Sharp students like you..... keep in touch.

[@rameshpal8937]

@@TIKLESACADEMY Thank You sir Videos are Soo helpful for me

[@vipulgupta6897]

You're great sir..I am watching all your videos continuesly and this is very helpful 😊

[@yashkumarkandoi8143]

Bhai EG lite! TRW padh

[@sandeeep1299]

Thankyou very much sir , aapka channel hamari bahut help kr rha h 🙏🙏🙏

[@TIKLESACADEMY]

Thanks for watching.

[@omkhare2837]

👨‍🏫🧑‍🎓 wonderful teacher

[@PANKAJMER-vd5lh]

Thank you for passing me in exams🎉

[@rajeevgupta6832]

thank you so much sir aapke video ke vajah se mere lines ke concept clear ho gye

[@TIKLESACADEMY]

Thanks for watching

[@nandeshkalshetti25]

Approx Answer: theta =45⁰ Phie=29⁰ Alpha=56⁰ Beta=45⁰

[@Mkargo]

Beta 43 yetoy maza

[@anirudhramesh6157]

correct

[@codingrecharge]

Sir, I was trying this drawing for 3 hours but didn't make out what to do, then I watched your video, and now I'm so much confident that I can solve any problem

[@samirpatel3677]

Thanks sir aa video joi ne mara doubt clear Thai gaya.👍🏾👍🏾👍🏾👍🏾👍🏾👍👍👍👍👍👍👍👍👍👍👍🎉🎉🎉🎉

[@ubro-rubicaseusa]

hii sam 8:18

[@apnvlogs7641]

hats off to you sir jai hind

[@SoniyaKulkarni-s2c]

Thank you so much sir for this college me toh kuch bhi samajh main nahi aaya tha aap ki wajah se mere 14 marks confirm ho gaye ❤❤❤❤❤❤❤❤❤❤❤❤❤

[@priyaranjansatpathy3346]

sir you are a life saviour for all of us😇😇

[@ManikantaCH-r1t]

A very big thanks to you sir😊🙏🏻🙏🏻Your explanations were just awesome and the thing i noticed is you revise every topics and rules whenever there's a time to🙏🏻🙏🏻🙏🏻🙏🏻... Thankyou so much sir🥰

[@girijeshsingh5923]

Theeta-45° Alpha -55° Phie -30° Beeta -45°

[@Darogaround3]

Sir i am from nepal i am watching your vedio very nice i like your vedios .....love you sir❣️❣️😍😘

[@unknown-dp5xq]

Always love you sir ❤

[@Himan_shu_044]

The end points of a line MN are, respectively, 30 mm and 70 mm in front of the VP while its HT is 10 mm in front of the VP. Draw the proje of line MN if it is located in the first quadrant and its front view is inclined at 30° to the HP. Its top view is inclined at 45 to the XY line the true length of the line and its inclination to the VP. Locate its VT.

[@vaibhavbijjala]

Thank you so much sir your all video classes helped me a lot

[@TIKLESACADEMY]

Thanks for watching.

[@Roshanmandal1990]

Good explanation sir 🎉

[@sanskaribalak9841]

Thank you sir your video is very helpful 🙏🙏🙏😁😁

[@swetachaudhary9220]

Thank you so much for help sir. We are very greatful.😊🙏

[@sayadk]

Thank you sir 😊

[@ManishThakur-qu5zn]

You are not a teacher you are god for me....😘💗

[@swaroopapte3440]

god of engg drawing

[@goldenkumar8387]

awesome teaching👍

[@sagarjaideepdeshmukh]

sirjee k charno mai koti koti pranam

[@TIKLESACADEMY]

Thanks for watching

[@mrprahlad3382]

Thanku so much sir ❤

[@rajatkarmakar7151]

Draw xy line and one projector. 2) Locate a’ 10 mm above xy line and a 15 mm below xy line. 3) Draw locus from these points. 4) Draw FV 50º to xy from a’ and mark b’ cutting 55 mm on it. 5) Similarly draw TV 60º to xy from a and drawing projector from b’. Locate point b and join ab. 6) Then rotating views as shown, locate True Lengths ab1 and a’b1’ and their angles with HP and VP.

[@rajatkarmakar7151]

Please sir solved

[@ansariazeem678]

Best teaching ever even maths also you teacha the best ever

[@PavanG-b2k]

thankyou soo much sir

[@anuragpal7829]

You are great sir for easy explain us

[@learnwithpujan3422]

Thank you so much sir 😊😊

[@anamay8503]

Sir I want to pay lot of thanks sir your videos is very helpfull to all engineering student 🙌🏼🙏🏼🙏🏼

[@pateljayantibhai245]

sir u r brilliant really thank u very much for sharing ur best knowledge.,,,,,,,,,,,,.........................

[@sumitkhadka3162]

Love you sir form Nepal ❣️

[@mahakalbhakt2020]

Thanks you very much sir ji aap bahut accha padhate hai❤❤❤❤

[@vishalkarmur3153]

Wonderful teacher sir 👍🏻💯

[@WAHEGURU806]

WAHEGURU JI THANKS SIR JI 🙏🙏🙏🙏🙏🙏🚩

[@Royal_Priyanshu]

Thank you sir 🥰🥰

[@ishinde4985]

Thanks Sir all Concepts are clear

[@nishchalacharya9789]

thank you sir......very helpful... video

[@lalithab2850]

Sir tq somuch for this video and I truly admire ur teaching 😊😊😊

[@kundanmali2405]

Thanks to you sir almost you clear all doubt of me

[@TIKLESACADEMY]

Thanks for watching

[@manishpal1501]

Love You sir for great help

[@adityarajsinghrathor]

Dhanyawad Guruji 🙏🙏🙏🙏🙏

[@saiwaydande8774]

Thanks sir🙏

[@chins379]

Sir,I am very thankful to you for these useful vedios, plz give us more guidance of different subjects Sir plz upload ,concept of HT and VT

[@TIKLESACADEMY]

Thanks for watching

[@polymers9987]

​@@TIKLESACADEMY sir in2022-2023 for 1 St year 2 sem we have got reduced the summ based on traces

[@mdtausifalam3750]

Aap bahut hi achha sikhate h

[@kinleytshering6535]

Thank you so much sir

[@Varma_Vlogs18]

Thank u so much Ur videos are so helpful

[@shwetapatil6223]

Thank you so much sir,,,,,,,mai iss subject se irritate ho gayi thi kuch samajh hii nahi aa raha tha.....you are amazing 🤩

[@yohohohoheyy]

Explained so easily, thanks sir

[@KrushnaBirajdar-de5ih]

Very nice

[@manishtrivedi3546]

Please give the example when Mid point of line in VP or HP

[@bijendrakumarprajapati3780]

Thanks 🌹🌹🌹🌹🌹sir

[@aadrshthakur7538]

Thanks 🙏 sir it helps us to lot

[@r4izyes]

Really awesome explanation sir. Thanks a lot

[@harshithsai704]

The top view of a line is 65mm long and is inclined at 300 to the reference line. One end is 20mm above HP and 10mm in front of VP. The other end is 60mm above HP and is infront of VP. Draw the projections and find the true length of the line and its true inclinations to HP and VP.

[@ssrockgaming3757]

Sir ji aapka concept achha hai. Samajh me bhi aata hai But sir mai hindi medium se hun Meri english kamjor hai

[@enakshisaha7495]

Thanku sir🙏🙏🙏🙏

[@hetpatel2992]

The Approximate values are : Theta = 45 degree Fi = 30 degree Alpha = 60 degree Beta = 50 degree

[@EDU_Chu]

Bro alpha is 55 and beta is 45

[@hitesh5783]

@@EDU_Chu yes right

[@a2zcricket876]

@@EDU_Chu correct

[@deep14054]

bro alpha = 55, beta = 45

[@carelessff7882]

Alpha = 55°

[@mallanagoudapatil-e3i]

SUPPER SIR

[@RahulKumar-nu8dk]

The top view of a 75 mm long line AB measures 65 mm, while the length of its front view is 50 mm. It’s one end A is in the H.P. and 12 mm in behind the V.P. Draw the projections of AB and determine its inclinations with the H.P. and the V.P.

[@chetanbarhate4840]

Best teacher

[@Mahadev.parvatiiji]

Thank so much Guru ji

[@zykieba]

The top view of a 75 mm long line measures 55 mm. the line is in the V.P., its one end being 25 mm above the H.P. draw its projections.

[@1anu_ra-dha.]

Thankyou so much sr 🙏😊👍

[@gouravpatil17]

osmmmmmm🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏❤

[@moanisanis7229]

Thanks sir ji

[@TIKLESACADEMY]

Thanks for watching

[@generalstudybydileepsir4914]

Thank you sir 🙏🙏🙏

[@salonipatelsp4042]

😊

[@rushabhparikh4937]

Sir really very thankyou

[@TIKLESACADEMY]

Thanks for watching

[@saiwaydande8774]

❤

[@prankrishnadas3383]

THANK YOU SO MUCH SIR❤❤

[@abhijitjadhav14489]

Theta =47 Alpha=56 Phi=29 Beta=45

[@shekharchauhan1107]

Sir aapka coat pant bahut badiya hai ❤❤

[@mdtausifalam3750]

Thank you so much sir

[@mR_Sachin51]

Theta - 49 Phi - 29 Alpha - 59 Beta - 47

[@ashishmane5773]

Very good sir

[@davdakunal7566]

❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

[@nirajanadhikari6255]

sir full help hjr ko veadio🇳🇵🇳🇵🇳🇵

[@lel3923]

Howdy neighbour🇮🇳

[@jagmohansaini1398]

Sir sction of a wall including foundation walk topic

[@TIKLESACADEMY]

Thanks for watching

[@khushikirnapure]

Thanks sir

[@Nabanyash_Sarmah1]

Sir❤️

[@pranjalchouhan5658]

The front view of a straight line CD is 50 mm long and is inclined at 60° to XY. The end C is 10 mm above HP and 20 mm in front of VP. Draw the projections of this line if it is inclined at 45° to VP. Determine the inclination of the line with HP

[@prakashrathour7324]

Jai Hind sir

[@SnehalKalbhar]

Sir please share it's answer

[@karanthakur11]

2:34 Given 7:20 draw

[@EngineeringDrawing-pv6qq]

Theta = 45 degree Alpha= 55 degree Phai = 30 degree Beta =47 degree

[@cm-b-45vipulkumarsharma58]

theta= 45 phi= 30 alpha= 55 beta= 45 fv= 70mm tv= 55mm

[@chiragr1609]

Yesss! Got the same answer!

[@EDU_Chu]

Correct

[@a2zcricket876]

👍👍👍

[@SreeramVin]

good problem , ig

[@Kunnu_bhaiya_]

Theta is 50°. And pai is 30°

[@pritmajmudar1647]

Sir jab point 1st and 3rd quadrant me ho,waisa ek line projection ka case solve kijiye

[@TIKLESACADEMY]

Yes off course.... please wait....more videos on Straight Line will come soon covering all quadrants.... keep in touch

[@pritmajmudar1647]

@@TIKLESACADEMY Thank you sir Your videos are really helpful for me

[@hariravi6409]

Help in class 11th engineering graphics paper

[@durgaparthamikshalaisbavip2557]

theta 46 phi 28 alpha 65 bita 43

[@looserwinner329]

Ok hai

[@navmithapuria3877]

Sir, i have a doubt Sir jb hum lines draw krte hain to tub hmari length question me given jo hoti ha utni hi aati hai

[@keval1282]

Theta = 45° and fai = 28°

[@foreverpandaxx]

getting same answer

[@keval1282]

@@foreverpandaxx ok

[@SanskarShinde-pm8rr]

Bahut tej ho reh ho ha

[@aniketkolte8958]

Tysm😊Sir For all varieties Of Problems On proj.of lines Theta:-48° Phi:-28° Alpha:-58° Beta:-44°