Find the Area of the Triangle | Quick Trick | 3 Easy Methods

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Your quick trick works so well my friend! Nice job once again haha!

drpkmath
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AC= AB*Cos15, CB= AB*Sin15
Area = ½AC*CB= ½*4*4*Sin15Cos15
=4*(2Sin15Cos15)=4*Sin30=2

harikatragadda
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AC = 4cos(15)
BC = 4sin(15)
Area = (1/2)4²cos(15)sin(15) = 8cos(15)sin(15) = 4sin(30) = 2

JSSTyger
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First method is good. But for 2nd 3rd method, Sin15 is not a known value. Better use known sin30=2sin15cos15 and solve.

hrisheekeshbhave
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Another way would be to use the Sine formula for triangles.
a/sin A = b/sin B = c/sin C.
CB/sin 15 = AC/sin 75 = AB/sin 90
So CB/0.259 = AC/0.966 = 4/1.000
CB/0.259 =4 and AC/0.966 = 4
Cross multiplying:-
CB = 4x0.259 = 1.036
AC = 4x 0.966 = 3.864

montynorth
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The ratio of the sides of a right triangle with three angles of 15, 75, and 90 degrees is sqrt(6)-sqrt(2) : sqrt(6)+sqrt(2) : 4, so the answer is easy to find by calculating Since it is in the form of a product of sum and difference, (6-2)/2=4/2=2. Another solution is to paste congruent triangles symmetrically to form isosceles triangles of 30 degrees, 75 degrees, and 75 degrees. If you draw a perpendicular line from the base angle, you will see triangles of 30 degrees, 60 degrees, and 90 degrees. In this case, we know that the hypotenuse is 4, so the adjoining side is 2. And the 2 on the adjacent side is the height if the hypotenuse is the base. This is the area of two congruent triangles stacked together, so divide by 2 and the answer is 2.

ebich
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I went for a*b/2 or (4*cos(15))*(4*sin(15))/2=2. (But I admit, I used a calculator… - but that’s what our professor did as well!) I am at odds with knowing the fractions of the trigonometric functions by heart. But the other solutions were interesting as well. Never heard of ab*sin(theta) in school - it‘s cool! And I found it cool that this allowed for a solution without a calculator!

philipkudrna
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Sir through your video I learned new new methods of mathematics which I didn't seen before

dhrubajyotisarma
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Area=½ab sinC=½.4.4cosC.sinC
=4.sin30°=2
(C=15°)

dhrubajyotidaityari
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thx so much sir i like this maths problems .I watch your videos everytime and i practice those problems before the correct

أستاذرضاعبدهالكحلاوي
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Thank you, Sir. Every time I watch your video I wanted to thank you a million times. Please Sir, allow me to ask this question. Why you chose CB as the base of the triangle? Why you did not choose AB or AC as the Base of the triangle? If my question is ignorant, please, forgive me.

michelg
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ಜೈ ಶ್ರೀ ರಾಮ್ 🙏THAN Q"PreMATH"for presenting super video of solution to AREA of ∆ USE FULL & interested MAY GOD BLESS U !!!

sreedharaks
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You can solve this by not being sin cos tan. Use 4 peace of this triangles like puzzle. You will find equilateral triangle. So (4×4/2)/2=2

Couch-Tomato
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I think an easier solution is:

cos 15 = h/4 => h = 4*(cos 15)
sin 15 = b/4 => b = 4*(sin 15)

Hence
A = (h*b)/2 = (4*(cos 15)*4*(sin 15))/2
= 8*(cos 15)*(sin 15)
= 8*(1/2)/2 ( because sin 30 = 1/2 )
= 2

clementyip
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Cool, nice simple one, somehow, compared to your other challenges, trig seems to be the easy way out. 😃 great example though 😎

theoyanto
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Consider triangle CB is opp and AC is adj area of triangle is 1/2(sin15*4+cos15*4)=2

rajendram
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The area is merely: 1/2 * base * height == 1/2 * (hypotenuse * sin x) * (hypotenuse * cos x) == 1/4 * (hypotenuse ** 2) * sin(2 * x)

RL-mvhj
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By seeing this video, for me those 3 methods are become easy sir .

India-jqpi
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You can make the 2nd method much cleaner:
sin 15 = (√3-1) / 2√2 so sin 15 = BC / AB = BC / 4 = (√3-1) / 2√2 which gives BC = 4 x (√3-1) / 2√2.
cos 15 = (√3+1) / 2√2 so cos 15 = AC / AB = AC / 4 = (√3+1) / 2√2 which gives AC = 4 x (√3+1) / 2√2.
Area ABC = 1/2 x BC x AC = 1/2 x 4 x (√3-1) / 2√2 x 4 x (√3+1) / 2√2 = 8 x (√3-1) / 2√2 x (√3+1) / 2√2x = 8 x (√3-1) x (√3+1) / 8 = (√3-1) x (√3+1) = 3 - 1 = 2.

easy_s
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Triangle ADB is an equilateral triangle. Then all sides must be equal to 4
But in second method the side BD is 1.036+1.036= 2.072
How?

chandrapatiramakoteswarara