You can use the arccosine formula. A =arc cos (b^2 +c^2 -a^2)/2bc , B =arc cos (a^2 +c^2 -b^2)/2ac , C =arc cos (a^2 +b^2 -c^2)/2ab Here, A is lower left apex, B is lower right apex, C is top apex. Side b here is 3, side a is 5, side c is 7. Angle at A is 38.213 degrees, Angle C is 120 degrees, Angle C is 21.787 degrees.
Great stuff folks.My solution unless otherwise posted os finding area using Herons formula and plugging into formula area equals 0.5x3x5sinC therefore sinC equals area÷0.5x5x3 equals 6.49319÷7.5 equals 0.866025 therfore C is 120d.HTH
Using *cosine rule* cos A = (3^2 + 5^2 - 7^2) / (2 x 3 x 5) = (9 + 25 - 49 = -15) / (2 x 15) = -1 / 2 = cos 120 (since 49 > 9 + 25 an obtuse angle was anyway expected). So, A = 120 deg.
Alternatively, you can draw a perpendicular line from A to meet BC to set up the equation using Pythagoras theorem now that the two right angled triangles have common height. Those who talk of cosine rule doesn't appreciate the beauty of maths, imo.
You can use the arccosine formula. A =arc cos (b^2 +c^2 -a^2)/2bc , B =arc cos (a^2 +c^2 -b^2)/2ac , C =arc cos (a^2 +b^2 -c^2)/2ab Here, A is lower left apex, B is lower right apex, C is top apex. Side b here is 3, side a is 5, side c is 7.
Angle at A is 38.213 degrees, Angle C is 120 degrees, Angle C is 21.787 degrees.
Coulda just used the law os cosines)
But elementary proof are nice😊
use cosine rule
Great stuff folks.My solution unless otherwise posted os finding area using Herons formula and plugging into formula area equals 0.5x3x5sinC therefore sinC equals area÷0.5x5x3 equals 6.49319÷7.5 equals 0.866025 therfore C is 120d.HTH
just use law of cosine
Excellent video man 😊 👍
it looks like 115 degrees. But I have a protractor. I measured it and it was 120 degrees.
nice vid keep up
Using *cosine rule* cos A = (3^2 + 5^2 - 7^2) / (2 x 3 x 5) = (9 + 25 - 49 = -15) / (2 x 15) = -1 / 2 = cos 120 (since 49 > 9 + 25 an obtuse angle was anyway expected).
So, A = 120 deg.
There is no necessity to find the measure of angle CAD
Angle BAC =Angle ACD + angle ADC = 30 +90=120 degrees
120 ^0 ?
Alternatively, you can draw a perpendicular line from A to meet BC to set up the equation using Pythagoras theorem now that the two right angled triangles have common height.
Those who talk of cosine rule doesn't appreciate the beauty of maths, imo.