# Solution of Triangle.

Figure: Standard notation for a triangle

A triangle consists of three sides and three angles. They are collectively known as the six elements of the triangle. To solve a triangle means to find the remaining elements when three other elements are known.

In solving a triangle, the following different cases arise.

1. Three angles given

2. Three sides given

3. Two angles and one side given

4. Two sides and the included angle given

5. Two sides and an angle opposite to one of them given

The right-angled and oblique triangles are also included in the above cases.

CASE I Three angles given

a:b:c = sinA : sinB : sinC

CASE II Three sides given

${\rm{cosA}} = \frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$

Case III Two angles and one side given

B =180o – (A+C)

To find the other two sides Band C, we use the formula

$\frac{{\rm{a}}}{{{\rm{sinA}}}} = \frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinc}}}}$

Case IV Two sides and the included angle given

a2= b2+ c2-2bc cosA

CASE V Two sides and an angle opposite to one of them given

Let b, c and B be the given parts. Then, from the formula we can find C.

$\frac{{\rm{b}}}{{{\rm{sinB}}}} = \frac{{\rm{c}}}{{{\rm{sinc}}}}$

Example1: The angles of a triangle are 1050 and 115o, find the ratio of its sides

Soln:

In ΔABCm Let A = 105°, B = 15° Then C = 180 – (A + B) = 180 – (105+15) = 60°.

Now, we have,

Or, CasinA=bsinB=csinC

Or, asin105∘=bsin15∘=csin60∘

Or, a3+122 = b3−122

Or, a3+1 =zb3−1 = √c6

So, a : b : c = (3√+1):(3√−1):6√

Example:2 Given, a = 2, b = 22, c =33 + 1.solve the triangle.

Soln:

Given, a = 2, b = 2√2, c =3√3 + 1.

To find: A, B, C.

Now,

cosA = b2+c2−a22bc = 2+(3+1)2−42.2.(3+1)

= 2+3+23+1−42.2.(3+1)

= 2+232.2(3+1)

= 2(1+3)2.2.(3+1)

= √12 = cos45°.

Again.

Cos B = c2+a2−b22ca = (3+1)2+4−22.(3+1).2

= 3+23+1+4−24.(3+1)

= 6+234.(3+1)

= 23(3+1)4(3+1)

= 3√232 = cos30°.

Example:3 If a = 2, b = 4, c = 60°, find A and B.

Soln:

Here, a = 2, b = 4, c = 60°.

To find : A and B.

We have, cosC = a2+b2−c22ab

Or, 12 = 4+16−c22.2.4 [ C = 60°]

Or, 8 = 20 – c2.

Or, c2 = 12

So, c = 23√3.

Now, asinA=bsinB=csinC

Or, 2sinA=4sinB=2332  [c = 23√23 and C = 60°]

Or, 2sinA = 4sinB = 4

Now, sinA = 12à A = 30° and sinB = 1 à B = 90°.

Hence, A = 30°, B = 90°.

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