Hello maths bugs(🐞) and hivers(🐝)
Well come to meta's math riddles.
Today I have come with an another intersting math's probelm with it's sweet solution.There is a ∆ABC whose two sides along with thr angle between are given and we need to find out CO which is perpendicular to side AC.It is also given that BO is also perpendicular to side AB.
We can solve it using advance property but I am not going to use it. I will keep it simple, so that everyone can understand. To go to the soluion we will do some construction. Let's produce sides BO and AC to point P as you can see in the figure below 👇
Now we have a new triangle called ∆APB whose one side os given and it's are 90°, 60° and 30°. So, this a right angle triangle and we know the ratio of its sides. You can details about it in my previous post. Now check the following figure to check what's the measurement of the three sides of right triangle ∆APB.
Now look at the larger ∆CPO. We can easily find three angle of it and measurement of side CP is (6+4) cm that equals to 10 cm. Hence the same way as we have done in the above figure we can find other sides of ∆CPO as it is also a right angle triangle( 📐). Check the following figure.👇
Hence, the value of CO is 10/√3 cm. If we were asked to find value of BO, it can be found subtracting PB from PO. So ** BO** = (20/√3 - 3√3) cm.
All the figure used here are made by me using Android application. Figure may not exact; just consider the values and fact given. If you find any silly mistake, excuse me for that. Though tried checking several times.
I hope you find today's problem and solution helpful.
Thank you so much for coming around.
Have a nice day
All is well
Regards: @meta007
Hive indeed is a platform that has it all. Thanks for sharing.
Yeah that's correct. It can be main stream media unlike conventional ones.
Thanks man for stopping by.
You must be in love with math😀 what a pure dedication.
Keep doing the good works!
!PIZZA
Yeah I do love it.... Otherwise who cares to use brains! 🤪😜
Thanks for visiting dear.
This make me to remember mathematics calculations years ago
Hehe... Thanks for coming around.
Once again, I am here for the geometry.
!discovery 37
I am glad you like it.
Thank you so much for your visit and support. Have a great day.
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