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OnlineEdumath

Year of Establishment
March, 2020

6
 Experienced Faculty

We mentor/educate/teach learners Mathematics online via Google Meet App. Our Certified, Resilient and Productive Mathematics Educators make teaching and learning Mathematics fun for learners. Communicate us to mentor your child/children to become a lover of Mathematics, and a Mathematician

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Our Updates

We are smart and clever!

By OnlineEdumath   |  5th November, 2024
“Mathematics is the most beautiful and most powerful creation of the human spirit.” Stefan Banach
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Welcome to Online Edumath

By OnlineEdumath   |  4th September, 2024
This is Online Edumath website created by Ogheneovo Daniel Ephivbotor. Online Edumath falls in EDUCATION line of business. You can visit us offline at our office located at : Lugbe, Federal...
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We are smart and clever!

By OnlineEdumath   |  3rd July, 2024
For enquiry on what we do and on our shared Mathematics Questions and Solution review, please communicate us via our WhatsApp contact: +2349076614992  We are kindly to respond to your question(s)/...
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Thank You, Sir Bill.

By OnlineEdumath   |  18th April, 2024
The Management and Educators of Online Edumath kindly appreciate your words of affirmation and encouragement, Sir Bill. We are grateful! We pledge to keep working hard as we mentor/educate our lea...
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We are smart and clever!

By OnlineEdumath   |  18th April, 2024
Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them be...
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Mr. Daniel (Mathematics Educator)

By OnlineEdumath   |  26th August, 2023
Mr. Daniel is one of our Educators, He is a certified, hardworking, dedicated and passionate Mathematics Educator that aim at taking the sincere, profound knowledge of Mathematics to the comfort of...
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Mathematics Question and Solution

By OnlineEdumath   |  12th February, 2025
Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a...
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Mathematics Question and Solution

By OnlineEdumath   |  11th February, 2025
Sir Mike Ambrose is the author of the question. (a) Area yellow exactly in decimal cm² is; Area triangle with height 25.2274259042 cm and base 14.5sin(½atan(20/21)) cm = ½*25.2274259042*14.5sin(½a...
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Mathematics teaching and learning

By OnlineEdumath   |  10th February, 2025
Area ABED is; Area trapezium with parallel sides (4√(2)sin15) cm and (4+4√(2)sin15) cm, and height 4√(3) cm. = ½*4√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 2√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 24 cm²...
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Mathematics Question and Solution

By OnlineEdumath   |  9th February, 2025
Blue area is; Area sector with radius 12 units and angle 30° - Area sector with radius 6 units and angle 60° - Area triangle with height 6 units and base 6sin120 units. = (30π*12²/360) - (60π*6²/3...
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Mathematics Question and Solution

By OnlineEdumath   |  8th February, 2025
Let the length of the ascribed regular hexagon be 2 units. Therefore it area will be; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. The green areas is; Area rectangle with...
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Mathematics Question and Solution

By OnlineEdumath   |  8th February, 2025
Let the side of the ascribed regular hexagon be 2 units. Area regular hexagon is; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. Area green is; Area kite with perpendicular lengths 4 un...
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Mathematics Question and Solution

By OnlineEdumath   |  7th February, 2025
a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Noti...
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Mathematics Question and Solution

By OnlineEdumath   |  7th February, 2025
tan15 = a/4 a = 1.07179676972 cm. Where a is AD. b² = 1.07179676972²+4² b = 4.14110472164 cm. Where b is DF. c = 4-1.07179676972 c = 2.92820323028 cm. Where c is CD. sin60 = d/2.92820323028  d =...
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Mathematics Question and Solution

By OnlineEdumath   |  6th February, 2025
a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin9...
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Mathematics Question and Solution

By OnlineEdumath   |  6th February, 2025
a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659...
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Mathematics Question and Solution

By OnlineEdumath   |  5th February, 2025
a = 15-11.25  a = 3.75 m. b = atan(3/3.75)° c = 90-atan(3/3.75) c = 51.34019174591°  Calculating Length CD. tan(51.34019174591) = 15/CD CD = 15/tan(51.34019174591)  CD = 12 m....
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Mathematics Question and Solution

By OnlineEdumath   |  5th February, 2025
Let the square side be a. tan60 = a/b b = a/tan60 cm. Where b is BS. Calculating a. 2b+a = 6 2(a/tan60)+a = 6 a((2/tan60)+1) = 6 a = 6/((2/tan60)+1) a = 6(2√(3)-3) cm. Area Shaded...
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Mathematics Question and Solution

By OnlineEdumath   |  4th February, 2025
a² = 10²+10²-2*10*10cos150 a = 19.31851652578 cm. Where a is AF which is equal AE. (19.31851652578/sin150) = (10/sinb) b = 15° Where b is angle BAF. tan15 = c/10 c = 2.67949192431 cm....
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Mathematics Question and Solution

By OnlineEdumath   |  4th February, 2025
a = atan(2)° b = 180-2a b = (180-2atan(2))° Where b is angle FBP. Notice; CQ = 5 cm. Calculating FP using sine rule. Let FP be c. (c/sin(180-2atan(2))) = (5/sin(atan(2))) c = 4...
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Mathematics Question and Solution

By OnlineEdumath   |  3rd February, 2025
Let the small inscribed square side be a. Let b be the side of the big inscribed square. It implies; b = 10-a Therefore; Calculating a. tan(90-75) = a/(10-a) a(1+tan15) = 10tan15...
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Mathematics Question and Solution

By OnlineEdumath   |  2nd February, 2025
Sir Mike Ambrose is the author of the question. Area R exactly in its simplest decimal form is; Area triangle with height 4.54406194214 units and base 2sin(atan(2)) units. = ½*4.54406194214*...
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Mathematics Question and Solution

By OnlineEdumath   |  1st February, 2025
Let the radius of the inscribed circle be r. The square side is 8 cm. b = 8-r c = 2+r Therefore; (2+r)² = 2²+(8-r)² 4+4r+r² = 4+64-16r+r² 4r = 64-16r 20r = 64 5r = 16 r = (16/5)...
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Mathematics Question and Solution

By OnlineEdumath   |  31st January, 2025
Sir Mike Ambrose is the author of the question. Let length AB = 2 units. Therefore; Side length of the square, x is; 2x² = 2² x = √(2) units. It implies; Area square is; x²...
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Mathematics Question and Solution

By OnlineEdumath   |  30th January, 2025
Sir Mike Ambrose is the author of the question. Inscribed square single side length is; 7.80540954246 cm Area green in cm² to 1 decimal place is; Area trapezoid with parallel sides 12 cm...
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Mathematics Question and Solution

By OnlineEdumath   |  29th January, 2025
Sir Mike Ambrose is the author of the question. Let the side let of the square be 2 units. Therefore; Area square is; 2² = 4 square units. It implies; Area R is; Area trapezoi...
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Mathematics Question and Solution

By OnlineEdumath   |  28th January, 2025
Area blue is; Area triangle with height 10 cm and base (10sin16.26025509) cm - Area semi circle with radius √(2) cm = (0.5*10*10sin16.26025509) - (½π√(2)²) = (14-π) cm²
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Mathematics Question and Solution

By OnlineEdumath   |  28th January, 2025
Let AB = x PB = x/4 Therefore; a = x - (x/4) a = (3x/4) units. b² = x²+(3x/4)² b = √(x²+(3x/4)²) units. c = (x-4) units. It implies that; ½(x*(3x/4)) + ½(4*√(x²+(3x/4)²)) + ½(4...
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Mathematics Question and Solution

By OnlineEdumath   |  27th January, 2025
Sir Mike Ambrose is the author of the question. Let the single side length of the square be 2 unit. Therefore; Area square is; 2² = 4 square units It implies; Area R is; Area...
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Mathematics Question and Solution

By OnlineEdumath   |  26th January, 2025
Red Area is; 7(area square with side 1 unit) - Area triangle with base 1 unit and height 4 units - Area square with side 1 unit - Area triangle with height and base 1 unit respectively. = 7(1...
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Mathematics Question and Solution

By OnlineEdumath   |  26th January, 2025
Cosine Rule Proof.
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Mathematics Question and Solution

By OnlineEdumath   |  25th January, 2025
The radius of the inscribed circle is 2 units. a = 2atan(⅓)° b = 2atan(⅕)° c = 180-a-b c = (180-2atan(⅓)-2atan(⅕))° sin(2atan(⅓)) = d/16 d = (48/5) units. cos(2atan(⅓)) = e/16 e =...
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Mathematics Question and Solution

By OnlineEdumath   |  24th January, 2025
8 = 4 a = (8-a) Cross multiply. 64-8a = 4a 64 = 12a a = ⅓(16) cm. b = 8-a b = ⅓(8) cm. c = atan(⅔)° d = atan(3/2)° ⅔ = 8/e e = 12 cm ⅔ = f/8 f = ⅓(16) cm. g = 8-f g = ⅓(...
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Mathematics Question and Solution

By OnlineEdumath   |  23rd January, 2025
Let the side of the green square be x. Therefore; x/(tan(atan¾)) + x + x/(tan(atan(4/3))) = 5 ⅓(4x) + x + ¼(3x) = 5 16x + 12x + 9x = 60 37x = 60 x = (60/37) units. It implies; Area square green...
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Mathematics Question and Solution

By OnlineEdumath   |  22nd January, 2025
Let AC be 1 unit. Area smaller pentagon is; ½(5)*(1/(2tan(36))) = 1.72047740059 square units. a² = 1²+0.5²-cos108 a = 1.24860602048 units. Where a is CD. (1.24860602048/sin108) =...
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Mathematics Question and Solution

By OnlineEdumath   |  21st January, 2025
a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103°  d = 60-b d = 22.41091053103° e...
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Mathematics Question and Solution

By OnlineEdumath   |  20th January, 2025
Sir Mike Ambrose is the author of the question. y = 2^(x) P coordinate is; P(1, 2) dy/dx = 2^(x)In2 For x = 1 dy/dx = 2In2 It implies; Q coordinate is; Q(0, (2-2In2)) Th...
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Mathematics Question and Solution

By OnlineEdumath   |  19th January, 2025
Let the two congruent lengths be 1 unit. tan40 = a/1 a = 0.83909963118 unit. cos40 = 1/b b = 1.30540728933 units. Calculating r, radius of the red inscribed circle. (r/tan(20)) + (r/t...
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Mathematics Question and Solution

By OnlineEdumath   |  18th January, 2025
Let the side of the regular pentagon be 1 unit. Calculating Area Red. a² = 1²+1²-2*1*1cos108 a = 1.61803398875 units. b = 108-36-60 b = 12° (c/sin12) = (1.61803398875/sin96) c = 0.3382...
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Mathematics Question and Solution

By OnlineEdumath   |  17th January, 2025
a² = 15²+12²-2*12*15cos120 a = 23.43074902772 cm. Where a is the side length of the equilateral triangle ABC. (23.43074902772/sin120) = (12/sinb) b = 26.32950349168° c = 60-b c = 33.67...
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Mathematics Question and Solution

By OnlineEdumath   |  16th January, 2025
Sir Mike Ambrose is the author of the question. Let the side length of triangle ABC be 1 unit. a = 120-78 a = 42° b = 60-42 b = 18° (1/sin42) = (c/sin18) c = 0.4618186516 unit. tan6...
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Mathematics Question and Solution

By OnlineEdumath   |  15th January, 2025
18² = 14²+16²-2*14*16cosa 448cosa = 14²+16²-18² a = acos((14²+16²-18²)/448) a = 73.39845040098° (18/sin73.39845040098) = (14/sinb) b = 48.18968510422° c = 180-48.18968510422-73.3984504009...
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Mathematics Question and Solution

By OnlineEdumath   |  15th January, 2025
Le the side of the pentagram be 1 unit. tan54 = a/0.5 a = 0.68819096024 units. cos54 = 0.5/b b = 0.85065080835 units. c = b+a c = 1.53884176859 units. Let the side length of the ascr...
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Mathematics Question and Solution

By OnlineEdumath   |  14th January, 2025
Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019...
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Mathematics Question and Solution

By OnlineEdumath   |  13th January, 2025
Let the side of the regular pentagon be 1 unit. Notice; The regular pentagon side is equal the regular heptagon side. a = ⅐(180*5) a = ⅐(900)°, the single interior angle of the regular he...
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Mathematics Question and Solution

By OnlineEdumath   |  12th January, 2025
a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376°  d = 0.5a d = 4.46056963484 cm....
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Mathematics Question and Solution

By OnlineEdumath   |  12th January, 2025
Area m : Area n is; (√(2)x)² : (½(3x))² = 2x² : ¼(9x²) = 4(2x²) : 4(¼(9x²)) = 8x² : 9x² = 8 : 9
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Mathematics Question and Solution

By OnlineEdumath   |  11th January, 2025
Area of coloured region is; Area triangle with height 53.3479501083 m and base 60sin(102.042575143) m + Area triangle with height 100.027406452 m and base √(5924)sin(12.5245961777) m = (½*53....
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Mathematics Question and Solution

By OnlineEdumath   |  11th January, 2025
Sir Mike Ambrose is the author of the question. Let beta be x. x = (½*atan(5/4)) Let alpha be y. y = (½*atan(4/5))° Therefore; Area blue exactly in cm² decimal is; Area triangle...
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Mathematics Question and Solution

By OnlineEdumath   |  10th January, 2025
a = 360-140-80 a = 140° cos20 = 1/b b = 1.06417777248 units. (1.06417777248/sin30) = (c/sin70) c = 2 units. d² = 4+1.064177772482²-4*1.06417777248cos140 d = 2.89711998506 units. (2.89711998506/...
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Our Media

media video Class with Tejiri (year 12) and Vwarhe (year 10)

Class with Tejiri (year 12) and Vwarhe (year 10)

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Vwarhe, js3 student (year 9).

Class with Vwarhe, js3 student (year 9).

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Joshua, year 1 pupil.

Class with Joshua, year 1 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Adesuwa, year 6 pupil.

Class with Adesuwa, year 6 pupil.

media video Class with Prince, year 5 pupil.

Class with Prince, year 5 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Class with Ellis, year 2 pupil.

Class with Ellis, year 2 pupil.

media video Mathematics and Science

Mathematics and Science

media video Identifying Prime Numbers

Identifying Prime Numbers

media video Quadratic Equation Using Formula

Quadratic Equation Using Formula

media video Introduction to Decimals

Introduction to Decimals

media video Constructing and Bisecting Lines and Angles

Constructing and Bisecting Lines and Angles

media video Math Story

Math Story

media video Mathematics is not difficult, it is a language

Mathematics is not difficult, it is a language

media video Comparing Numbers

Comparing Numbers

media video Place Value

Place Value

media video Telling Time

Telling Time

media video Young Sheldon

Young Sheldon

media video Young Sheldon

Young Sheldon

media video Young Sheldon

Young Sheldon

media video Student Bullied For Being Dumb, Turns Out He's Genius Neurosurgeon

Student Bullied For Being Dumb, Turns Out He's Genius Neurosurgeon

media video Maths at The Mela

Maths at The Mela

media video Math Story

Math Story

media video The Map of Mathematics

The Map of Mathematics

Learner's Feedback/Testimonial


Sir, my first term Mathematics result, I scored 99% and got first position

Ogheneruno

Ogheneruno

Online Edumath

Learner's Feedback/Testimonial


Sir, I was the third best pupil in class out of 30 pupils at the end of Autumn assessment

Prince

Prince

Online Edumath

Learner's Feedback/Testimonial


I was awarded Century Champion Award certificate as a celebrated Century Champion for completing my Century Work in Mathematics in my School

Oghosa

Oghosa

Online Edumath

Learner's Feedback/Testimonial


My Math Teacher in school do sometimes administer year 6 Math assessment to me, a year 5 pupil, and I often score 100%

Adesuwa

Adesuwa

Online Edumath

Learner's Feedback/Testimonial

I got Bronze Certificate representing my School in the junior (year 7 and 8) Mathematics Challenge Competition, 2023 organized by United Kingdom Mathematics Trust. Oghosa is in year 7

Oghosa

Oghosa

Online Edumath

Learner's Feedback/Testimonial


Sir, I got A+, the highest score in my final summer Mathematics assessment score in my class.

Adesuwa

Adesuwa

Online Edumath

Learner's Feedback/Testimonial

I scored 92% in Mathematics in my third term final Mathematics Examination assessment, Sir.

Ogheneruemu

Ogheneruemu

Online Edumath

Learner's Feedback/Testimonial

Sir, I got A+, the highest score in my final summer Mathematics assessment score in my class.

Oghosa

Oghosa

Online Edumath

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