Triangle call a geometrical figure with three parties and three corners. Finding of all these six elements of a triangle is one of problems of mathematics. If lengths of the parties of a triangle are known, then by means of trigonometrical functions it is possible to calculate corners between the parties.
It is required to you
- basic knowledge of trigonometry
Instruction
1. Let the triangle with the parties of a, b and page be set. At the same time the sum of lengths of two any parties of a triangle has to be more than length of the third party, that is a+b> c, b+c> an and a+c> b. Also it is necessary to find a-degree measure of all corners of this triangle. Let the corner between the parties of an and b be designated as α, a corner between b and c as β, and a corner between c and an as γ.
2. The theorem of cosines sounds so: the square of length of the party of a triangle is equal to the sum of squares of two other lengths of its parties minus the doubled work of these lengths of the parties on a cosine of the angle between them. That is make three equalities: a²=b²+c²−2×b×c×cos(β); b²=a²+c²−2×a×c×cos(γ); with²=a²+b²−2×a×b×cos(α).
3. From the received equalities express cosines of corners: cos(β)= (b²+c²−a²) ÷(2×b×c); cos(γ)= (a²+c²−b²) ÷(2×a×c); cos(α)= (a²+b²−c²) ÷(2×a×b). Now, when cosines of corners of a triangle are known to find corners use Bradis's tables or take arccosines from these expressions: β=arccos (cos(β)); γ=arccos (cos(γ)); α=arccos (cos(α)).
4. For example, let a=3, b=7, c=6. Then cos(α)= (3²+7²−6²) ÷(2×3×7)=11/21 and α ≈ 58.4 °; cos(β)= (7²+6²−3²) ÷(2×7×6)=19/21 and β ≈ 25.2 °; cos(γ)= (3²+6²−7²) ÷(2×3×6)=-1/9 and γ ≈ 96.4 °.
5. The same problem can be solved in a different way through the area of a triangle. At first find poluperimetr a triangle on a formula of p=(a+b+c) ÷2. Then count the area of a triangle on Heron's formula S= √ (p×(p−a) × (p−b) × (p−c)), that is the area of a triangle is equal to a square root from the work of a poluperimetr of a triangle and the differences of a poluperimetr and each of the parties of a triangle.
6. On the other hand, the area of a triangle is equal to a half of the work of lengths of two parties on a sine of the angle between them. S=0.5×a×b×sin(α) =0.5×b×c×sin(β) =0.5×a×c×sin(γ) turns out. Now from this formula express sine of corners and substitute the value of the area of a triangle received in the 5th step: sin(α) =2×S÷(a×b); sin(β) =2×S÷(b×c); sin(γ) =2×S÷(a×c). Thus, knowing sine of corners to find a-degree measure, use Bradis's tables or count arcsines of these expressions: β=arccsin (sin(β)); γ=arcsin (sin(γ)); α=arcsin (sin(α)).
7. For example, let the same triangle with the parties of a=3, b=7, c=6 is given. Poluperimetr is equal p=(3+7+6)÷2=8, S= Square √ (8×(8−3) × (8−7) × (8−6))=4√5. Then sin(α) =2×4√5÷ (3×7) =8√5/21 and α ≈ 58.4 °; sin(β) =2×4√5÷ (7×6) =4√5/21 and β ≈ 25.2 °; sin(γ) =2×4√5÷ (3×6) =4√5/9 and γ ≈ 96.4 °.