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Solution of Triangles

Sine Rule, Cosine Rule, Tangent Rule or Napier’s Analogy, Projection Rule, Trigonometric Ratios of Half Angles, Area of a Triangle, m-n Theorem, Apollonius Theorem

In any ∆ABC, the lengths of the sides opposite to angles A, B and C are denoted by a, b and c respectively. This means, BC = a; AC = b and AB = c. • Sine Rule

In any ∆ABC:

$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$, where k is a constant

• Cosine Rule

In any ∆ABC:

$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}$

$\mathrm{cos}B=\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}$

$\mathrm{cos}C=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$

• Tangent Rule or Napier’s Analogy

In any ∆ABC:

$\mathrm{tan}\frac{B-C}{2}=\frac{b-c}{b+c}\mathrm{cot}\frac{A}{2}$

• Projection Rule

In any ∆ABC:

$a=b\mathrm{cos}C+c\mathrm{cos}B$

$b=c\mathrm{cos}A+a\mathrm{cos}C$

$c=a\mathrm{cos}B+b\mathrm{cos}A$

• Trigonometric Ratios of Half Angles

In any ∆ABC:

• Area of a Triangle

If the area of the triangle is denoted by , then

$∆=\frac{1}{2}bc\mathrm{sin}A=\frac{1}{2}ca\mathrm{sin}B=\frac{1}{2}ab\mathrm{sin}C=\frac{abc}{4R}=rs=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Here, R is the radius of circumcircle of ∆ABC; r is the radius of inscribed circle of ∆ABC and 2s = a + b + c.

• m-n Theorem

Let D be a point on side BC of ∆ABC such that it divides the side BC in the ratio m : n. If and $\angle \mathrm{ADC}=\theta$, then

1. $\left(m+n\right)\mathrm{cot}\theta =m\mathrm{cot}\alpha -n\mathrm{cot}\beta$

2. $\left(m+n\right)\mathrm{cot}\theta =n\mathrm{cot}B-m\mathrm{cot}C$

• Apollonius Theorem

In $∆$ABC, if AD is the median through A, then ${\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}=2\left({\mathrm{AD}}^{2}+{\mathrm{BD}}^{2}\right)$.

Reality check
Question:1

What is the perimeter of the triangle whose lengths of sides are three consecutive natural numbers and the largest angle is twice the smallest angle?

A)
B)
C)
D)

Let the sides of the triangle be x, x + 1 and x + 2, where x is a natural number. Suppose ∠C = α is the smallest angle. ∴ ∠A = 2α

Applying sine rule in ΔABC, we have    …(1)

Also,      …(2)

From (1) and (2), we have    x = 4    (x is a natural number)

∴  x + 1 = 5 and x + 2 = 6

Perimeter of the triangle = x + (x + 1) + (x + 2) = 15 units

Hence, the correct answer is option C.

Question:2

The sides of a triangle are in arithmetic progression (A.P.). If the smallest angle of the triangle is θ and its largest angle exceeds the smallest angle by α, then what is the value of  $\mathrm{tan}\left(\theta +\frac{\alpha }{2}\right)$ ?

A)
B)
C)
D) Let the sides of ΔABC be ad, a and a + d, where d > 0.

The greatest side, AB = a + d and the smallest side, BC = ad.

So, ∠A is the smallest angle …

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