Cosine rule/Triangle Area and example problems
For all the formulae and examples below, 'a' is the side opposite angle A, i.e. BC.
Similarly 'b' = AC and 'c' = AB.
$$
\cos A = \frac{b^2 + c^2 - a^2}{2\,b\,c}
$$
$$ \cos B = \frac{a^2 + c^2 - b^2}{2\,a\,c} $$
$$ \cos C = \frac{a^2 + b^2 - c^2}{2\,a\,b} $$
Example Problems:
1. If $$ a = \sqrt{3},\quad b = \frac12\bigl(\sqrt{6} + \sqrt{2}\bigr),\quad c = \sqrt{2}, $$ then $$ \angle A =\;? $$
Answer: $$ \angle A = 60^\circ$$
Hint: Use Cosine rule
2. If (a+b+c)(a-b+c) = 3ac then $$\angle B = ?$$
Answer: $$60^\circ$$
Hint: Cosine rule
3. If $$ b = 3,\quad c = 4,\quad \angle B = 60^\circ $$ then how many such triangles are possible?
Answer: 0
Solution hint:
1. If you use Cosine rule for CosB, you will get a quadratic with 0 solutions.
2. If you use Sine rule, then you will get SinC > 1.
4. $$ \angle B = 30^\circ, c = b\sqrt3, \angle A = ?$$
Answer: $$ \angle A = 90^\circ$$
Hint: Use Sine rule.
5. Prove that $$ \frac{\cos A}{a} \;+\;\frac{\cos B}{b} \;+\;\frac{\cos C}{c} \;=\; \frac{a^2 + b^2 + c^2}{2\,a\,b\,c} $$
Solution hint: Simple application of Cosine Rule.
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Area of Triangle using 2 side lengths and included angles.
In a triangle ABC, if I drop altitude AD from A on BC, then $$ Area = \frac{1}{2}(ab\sin C) = \frac{1}{2}(ac\sin B) = \frac{1}{2}(bc\sin A) = \frac{abc}{4R} $$
Where 'R' = Circumradius(radius of Circumcircle which passes through all vertices.
Other area formulae: 1. Heron's formula using semiperimeter.
2. Using inradius A = s.r where 's' is semiperimeter and 'r' is inradius
Example question:
In a scalene triangle, 2 sides x,y are 60,63.
If x + 2A/x = y + 2A/y then find the largest side of the triangle.
A is the area of the triangle.
Solution:
Isolate A = (xy)/2 => right angle triangle. Then find the hypotenuse. Answer: 87
Example 2:
If sides of a triangle have this ratio => a : b : c = 4 : 5 : 6 then find R : r. R is circumradius and r is inradius.
Solution hint: Use 3 triangle area formulae -> Heron's, Circumradius and 3 sides, inradius and semiperimeter.
Answer: 16:7
$$ \cos B = \frac{a^2 + c^2 - b^2}{2\,a\,c} $$
$$ \cos C = \frac{a^2 + b^2 - c^2}{2\,a\,b} $$
Example Problems:
1. If $$ a = \sqrt{3},\quad b = \frac12\bigl(\sqrt{6} + \sqrt{2}\bigr),\quad c = \sqrt{2}, $$ then $$ \angle A =\;? $$
Answer: $$ \angle A = 60^\circ$$
Hint: Use Cosine rule
2. If (a+b+c)(a-b+c) = 3ac then $$\angle B = ?$$
Answer: $$60^\circ$$
Hint: Cosine rule
3. If $$ b = 3,\quad c = 4,\quad \angle B = 60^\circ $$ then how many such triangles are possible?
Answer: 0
Solution hint:
1. If you use Cosine rule for CosB, you will get a quadratic with 0 solutions.
2. If you use Sine rule, then you will get SinC > 1.
4. $$ \angle B = 30^\circ, c = b\sqrt3, \angle A = ?$$
Answer: $$ \angle A = 90^\circ$$
Hint: Use Sine rule.
5. Prove that $$ \frac{\cos A}{a} \;+\;\frac{\cos B}{b} \;+\;\frac{\cos C}{c} \;=\; \frac{a^2 + b^2 + c^2}{2\,a\,b\,c} $$
Solution hint: Simple application of Cosine Rule.
--------------------------
Area of Triangle using 2 side lengths and included angles.
In a triangle ABC, if I drop altitude AD from A on BC, then $$ Area = \frac{1}{2}(ab\sin C) = \frac{1}{2}(ac\sin B) = \frac{1}{2}(bc\sin A) = \frac{abc}{4R} $$
Where 'R' = Circumradius(radius of Circumcircle which passes through all vertices.
Other area formulae: 1. Heron's formula using semiperimeter.
2. Using inradius A = s.r where 's' is semiperimeter and 'r' is inradius
Example question:
In a scalene triangle, 2 sides x,y are 60,63.
If x + 2A/x = y + 2A/y then find the largest side of the triangle.
A is the area of the triangle.
Solution:
Isolate A = (xy)/2 => right angle triangle. Then find the hypotenuse. Answer: 87
Example 2:
If sides of a triangle have this ratio => a : b : c = 4 : 5 : 6 then find R : r. R is circumradius and r is inradius.
Solution hint: Use 3 triangle area formulae -> Heron's, Circumradius and 3 sides, inradius and semiperimeter.
Answer: 16:7
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