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

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THIS IS THE 6TH VIDEO ON "PROJECTION OF STRAIGHT LINE". TODAY WE WILL STUDY 4TH UNIVERSITY PROBLEM OF PROJECTION OF STRAIGHT LINE.
QUE - THE TOP VIEW OF 80mm LONG LINE AB MEASURES 55mm WHILE THE LENGTH OF IT'S FRONT VIEW IS 70mm. IT'S ONE END IS 15mm ABOVE THE HP AND 15mm INFRONT OF VP. DRAW IT'S PROJECTION & DETERMINE IT'S INCLINATION WITH HP and VP.
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TO WATCH ALL PREVIOUS VIDEOS FOLLOW THE LINK OF PLAYLIST GIVEN BELOW :
FOR ALL VIDEOS ON "PROJECTION OF POINTS",
• PROJECTION OF POINTS
FOR ALL VIDEOS ON "PROJECTION OF STRAIGHT LINE",
• PROJECTION OF STRAIGHT...
FOR ALL VIDEOS ON "PLAIN SCALE",
• PLAIN SCALE
FOR ALL VIDEOS ON "DIAGONAL SCALE",
• DIAGONAL SCALE
FOR ALL VIDEOS ON "ISOMETRIC PROJECTION",
• ISOMETRIC PROJECTION
FOR ALL VIDEOS ON "ORTHOGRAPHIC PROJECTION",
• ORTHOGRAPHIC PROJECTION
FOR ALL VIDEOS ON "ELLIPSE",
• ELLIPSE
FOR ALL VIDEOS ON "PARABOLA",
• PARABOLA
FOR ALL VIDEOS ON "HYPERBOLA",
• HYPERBOLA
FOR ALL VIDEOS ON "CYCLOID",
• CYCLOID
FOR ALL VIDEOS ON "EPICYCLOID",
• EPICYCLOID
FOR ALL VIDEOS ON "HYPOCYCLOID",
• HYPOCYCLOID
FOR ALL VIDEOS ON "INVOLUTE",
• INVOLUTE (ENGINEERING ...
FOR ALL VIDEOS ON "ARCHIMEDEAN SPIRAL",
• ARCHIMEDEAN SPIRAL
FOR ALL VIDEOS ON "TANGENT AND NORMAL",
• TANGENT AND NORMAL
#projectionofstraightline
#projectionofpoints
#engineeringdrawing
#engineeringgraphics
#tiklesacademy
#isometricprojection
#orthographicprojection

[@venkatrr8050]

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

[@girijeshsingh5923]

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

[@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

[@sameeksha1028]

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

[@vipulgupta6897]

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

[@yashkumarkandoi8143]

Bhai EG lite! TRW padh

[@rajeevgupta6832]

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

[@TIKLESACADEMY]

Thanks for watching

[@sy_25_nandeshkalashetti95]

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

[@Mkargo]

Beta 43 yetoy maza

[@anirudhramesh6157]

correct

[@sandeeep1299]

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

[@TIKLESACADEMY]

Thanks for watching.

[@ManishThakur-qu5zn]

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

[@swaroopapte3440]

god of engg drawing

[@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

[@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

[@sagarjaideepdeshmukh]

sirjee k charno mai koti koti pranam

[@TIKLESACADEMY]

Thanks for watching

[@samirpatel3677]

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

[@user-vo2nn9xt7o]

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🥰

[@vaibhavbijjala]

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

[@TIKLESACADEMY]

Thanks for watching.

[@omkhare2837]

👨‍🏫🧑‍🎓 wonderful teacher

[@xipingpungi2936]

Thank you very much sir

[@gauravnalavade1511]

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

[@swetachaudhary9220]

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

[@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

[@anamay8503]

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

[@sayadk]

Thank you sir 😊

[@WAHEGURU806]

WAHEGURU JI THANKS SIR JI 🙏🙏🙏🙏🙏🙏🚩

[@pateljayantibhai245]

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

[@sanskaribalak9841]

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

[@user-ef1ee4sy2x]

thankyou soo much sir

[@kundanmali2405]

Thanks to you sir almost you clear all doubt of me

[@TIKLESACADEMY]

Thanks for watching

[@DarogaRound3]

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

[@ansariazeem678]

Best teaching ever even maths also you teacha the best ever

[@lalithab2850]

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

[@manishpal1501]

Love You sir for great help

[@manishtrivedi3546]

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

[@KrushnaBirajdar-de5ih]

Very nice

[@unknown-dp5xq]

Always love you sir ❤

[@goldenkumar8387]

awesome teaching👍

[@ishinde4985]

Thanks Sir all Concepts are clear

[@Varma_Vlogs18]

Thank u so much
Ur videos are so helpful

[@adityarajsinghrathor]

Dhanyawad Guruji 🙏🙏🙏🙏🙏

[@moanisanis7229]

Thanks sir ji

[@TIKLESACADEMY]

Thanks for watching

[@nishchalacharya9789]

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

[@vishalkarmur3153]

Wonderful teacher sir 👍🏻💯

[@user-hu4cl8kt4s]

SUPPER SIR

[@bijendrakumarprajapati3780]

Thanks 🌹🌹🌹🌹🌹sir

[@salonipatelsp4042]

😊

[@davdakunal7566]

❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

[@shwetapatil6223]

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

[@anuragpal7829]

You are great sir for easy explain us

[@sumitkhadka3162]

Love you sir form Nepal ❣️

[@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.

[@rushabhparikh4937]

Sir really very thankyou

[@TIKLESACADEMY]

Thanks for watching

[@learnwithpujan3422]

Thank you so much sir 😊😊

[@yohohohoheyy]

Explained so easily, thanks sir

[@enakshisaha7495]

Thanku sir🙏🙏🙏🙏

[@ssrockgaming3757]

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

[@mdtausifalam3750]

Aap bahut hi achha sikhate h

[@Roshanmandal1990]

Good explanation 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.

[@mR_Sachin51]

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

[@Mahadev.parvatiiji]

Thank so much Guru ji

[@chetanbarhate4840]

Best teacher

[@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°

[@aadrshthakur7538]

Thanks 🙏 sir it helps us to lot

[@generalstudybydileepsir4914]

Thank you sir 🙏🙏🙏

[@r4izyes]

Really awesome explanation sir. Thanks a lot

[@khushikirnapure]

Thanks sir

[@prankrishnadas3383]

THANK YOU SO MUCH SIR❤❤

[@jagmohansaini1398]

Sir sction of a wall including foundation walk topic

[@TIKLESACADEMY]

Thanks for watching

[@nirajanadhikari6255]

sir full help hjr ko veadio🇳🇵🇳🇵🇳🇵

[@lel3923]

Howdy neighbour🇮🇳

[@1anu_ra-dha.]

Thankyou so much sr 🙏😊👍

[@shekharchauhan1107]

Sir aapka coat pant bahut badiya hai ❤❤

[@gouravpatil17]

osmmmmmm🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏❤

[@ashishmane5773]

Very good sir

[@yashwanth4358]

Sir u r not explaining about initial front view,top view and final front view,top view will u plz teach us

[@Nabanyash_Sarmah1]

Sir❤️

[@EngineeringDrawing-pv6qq]

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

[@twblogger7243]

Sir questions me TV 68 mm hoga nahi to diagram galat ho jaega
Please help sir 🙏🙏
(Is it right or wrong...???)

[@aniketkolte8958]

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

[@prakashrathour7324]

Jai Hind sir

[@SreeramVin]

good problem , ig

[@Kunnu_bhaiya_]

Theta is 50°. And pai is 30°

[@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]

👍👍👍

[@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

[@222chaitaliramanwad7]

Tq

[@TIKLESACADEMY]

Thanks for watching

[@ericvaish]

2:40

[@vijaykumarprasad2615]

Isometric jo ellipse ke sath ho uska view bataiye sir

[@aslampathan2321]

Sir I have watched all 4 type of problem video but all are in first quadrant where are 2,3,4 quadrant video tommorow is my exam sir plz help

[@TIKLESACADEMY]

Thanks for watching all my videos. Problems on Straight Line lying in 2nd, 3rd & 4th quadrants will come soon....
anyway...
BEST OF LUCK FOR THE EXAMS.

[@Rajtvno1]

Very important question

[@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

[@VishalYadav-bh6sy]

Love you sir

[@TIKLESACADEMY]

Thanks for watching

[@keval1282]

Theta = 45° and fai = 28°

[@foreverpandaxx]

getting same
answer

[@keval1282]

@@foreverpandaxx ok

[@SanskarShinde-pm8rr]

Bahut tej ho reh ho ha

[@durgaparthamikshalaisbavip2557]

theta 46
phi 28
alpha 65
bita 43

[@looserwinner329]

Ok hai

[@Mcg-vn8xd]

Theta 47 degree
Fai 30 degree

[@pushpendraprajapati8065]

Namaste sir

[@karanthakur11]

2:34 Given
7:20 draw

[@Naveen.1116]

Sir kya true length kbhi bhi same nhi hote
Sirf top view and front view hi same hote hai??

[@pranshusharma3653]

Sir or anyone can tell that in this series has sir taught about rotational and trapezoidal methods?
Pls reply

[@SCSGamer00YT]

sir question mai bola gya hai top view of 80mm long line iska matlb jo hmara top view hai tv uski hme true line de rhi rhi hai to wo hme niche draw krna chahiye tha na but aapne use fv par draw liya kyu sir ?

[@YashPatel-zl3ex]

True length is equal for top view and front view
80mm is true length of line so it can drawn at both views

[@idiotme5478]

Exactly also waha par 55 mm ka DBET diya hain woh tv thodi hain

[@idiotme5478]

​@@YashPatel-zl3extht we know but jab quest mein TL ka koi mention hi nhi toh kaha se aaya TL

[@bandinipatil41]

α=56°
β=45°
Theta=47°
Φ=29°