Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Get extra Rs. 95 off
USE CODE: COUPON10

```Solution of Triangles

In a triangle ABC, the vertices and the angles are denoted by capital letters and the sides by small letters.

In the figure given below, the sides opposite to angles A, B, C are denoted by a, b, c respectively. The area of the triangle is denoted by S or Δ.

Basic Formulae and Results:

Some of the basic trigonometry formulae depicting the relationship between the sides and angles of a triangle are listed below:

Sine rule

This rule states that the sides of any triangle are proportional to the sines of the angle opposite to them in ?ABC, i.e.

a/ sin A = b/ sin B = c/ sin C

Note:

The rule can further be expressed as (sin A)/a = (sin B)/b = (sin C)/c

The rule is extremely useful for expressing the sides of a triangle in terms of sines of angle or vice versa as per the requirement of the question. Hence, for the form a/ sin A = b/ sin B = c/ sin C = k, we have,

a = k sin A, b = k sin B, c = k sin C

and for (sin A)/a = (sin B)/b = (sin C)/c = l, we have

sin A = la, sin B = lb, sin C = lc

Cosine rule

In any triangle ABC, the following results hold good:

cos A = (b2 + c2 - a2 )/2ac

cos B = (a2 + c2 - b2)/2ac

cos C = (a2 + b2 - c2)/2ab

Note: In case

∠A = 60°, then b2 + c2 - a2 = bc

∠B = 60°, then a2 + c2 - b2 = ac

∠C = 60°, then a2 + b2 - c2 = ab

Trigonometric ratios of half-angles:

sin A/2 = √(s-b) (s-c)/bc

sin B/2 = √(s-b) (s-c)/bc

sin C/2 = √(s-a) (s-b)/ab

cos A/2 = √s(s - a)/bc

cos B/2 = √s(s - b)/ca

cos C/2 = √s(s - c)/ab

tan A/2 = √(s - b) (s - c)/s(s - a)

tan B/2 = √(s - c) (s - a)/s(s - b)

tan C/2 = √(s - a) (s - b)/s(s - c)

Projection rule:

a = b cosC + c cosB

b = c cosA + a cosC

c = a cosB + b cosA

Semi-perimeter of the triangle

If ‘s’ is assumed to be the perimeter of the triangle then s = a + b + c / 2

Area of a triangle

If Δ is the area of the triangle ABC then

Δ = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin C

Δ = √s(s - a) (s - b) (s - c), this is also called as the Hero’s Formula

Δ = (a2 sin B sin C)/ 2 sin (B + C) = (b2 sin C sin A)/ 2 sin (A + C) = (c2 sin A sin B)/ 2 sin (A + B)

Some illustrations based on these formulae

Illustration:

Find out the length of x in the figure given below

Solution:

Whenever we need to find the length of a side, a different sort of sine formula is used which simplifies the calculation of length:

Using the result, a/sin A = b/ sin B

Here, a = 7, b = x, A = 60°, B = 80°

7/ sin (60°)) = x / sin (80°)

Multiplying both sides by sin (80°) for simplification we get

(7/sin (60°)) sin (80°) = x

Hence, on simplification we get

x = 7.96, hence this is the required length.

Illustration:

Find the missing side in the diagram below:

Solution:

Here, we will have to apply the cosine formula.
a2 = b2 + c2 - 2bc cos A
Hence substituting the values we get,
h2 = 882 + 1462 – 2×88×146×cos 53°
h2 = 13595.761.

For more clarification, refer the video

Some vital trigonometry theorems associated with solution of triangles:

Napier’s analogy

In any triangle ABC, the following results hold good:

tan (B – C)/2 = (b – c)/(b +c) cot A/2
tan (C – A)/2 = (c –a)/(c + a) cot B/2
tan (A – B)/2 = (a – b) /(a + b) cot c/2

m-n theorem

If D be the point on the side BC of a triangle ABC which divides the side BC in the ratio m: n, then with respect to the figure given below, we have:

(m + n) cot θ = m cot α – n cot ß.

(m + n) cot θ = n cot B – m cot C.

Apollonius theorem

Consider a triangle ABC. As depicted in the figure given below, D is the median through A.

Then AB2 + AC2 = 2(AD2 + BD2).

Process of Solution of Triangles:

A triangle is known completely if the three sides and angles are known. These constitute the elements of the triangle. Thus, if we have any three elements of a triangle (other than the three sides) say two sides and the included angle or alike then we can easily find the rest of the elements using the various formulae. Hence the triangle is known completely and this process is called the solution of triangles.

We list below the various ways of preceding a given problem when some elements are unknown:

If the three sides a, b, c are given, angle A is obtained from

tan A/2 = √(s - b) (s - c)/s(s - a) or cos A = b2 + c2 - a2 / 2bc.

B and C can be obtained in the similar way.

The cosine formula should be used only when a, b and c are small numbers.

If two sides b and c and the included angle A are given, then

tan (B – C)/2 = (b – c)/ (b + c) cot A/2 gives (B – C)/2.

Also B + C/2 = 90o - A/2, so that B and C can be evaluated. The third side is given by a = b sin A/sin B or a2 = b2 + c2 – 2bc cos A.

If two sides b and c and the angle B (opposite to side b) are given, then sin C = c/b  sin B, A = 180o – (B + C) and b = b sin A/sin B give the remaining elements.

If b < c sin B, there is no triangle possible (Fig. 1)

If b = c sin B and B is an acute angle, then only one triangle is possible (Fig. 2)

If c sin B < b < c and B is an acute angle, then there are two values of angle C (Fig. 3).

If c < b and B is an acute angle, then there is only one triangle (Fig. 4).

This is, sometimes, called an ambiguous case.

Alternative Method:

By applying cosine rule, we have cos B = a2 + c2 - b2/2ac

⇒ a2 – (2c cos B) a + (c2 – b2) = 0

⇒ a = c cos B + √(C cos B)2 - (c2 - b2)

⇒ a = c cos B + √b2 - (c sin B)2.

This equation leads to the following cases:

Case –I

If b < c sin B, no triangle is possible.

Case –II

Let b = c sin B. There are further the following cases:

(a) B is an obtuse angle

⇒ cos B is negative. There exists no such triangle.

(b) B is an acute angle

⇒ cos B is positive. There exists only one triangle.

Case –III

Let b < c sin B. We have the following cases:

B is an acute angle ⇒ cos B is positive. In this case two values exist if and only if c cos B > √b2 - (c sin B)2 or

c > b ⇒ Two triangles are possible.

If c < b, only one triangle is possible.

B is an obtuse angle ⇒ cos B is negative. In this case triangle will exist if and only if √b2 - (c sin B)2 > c cos B ⇒ b > c. So in this case only one triangle is possible.

If b < c there exists no triangle.

If one side a and angles B and C are given, then A = 180o – (B + C), and b = a sin B/sin A, c = a sin C/sin A .

If the three angles A, B, C are given, we can only find the ratios of the sides a, b, c by using the sine rule (since there are infinite similar triangles possible).

You might like to refer some of the related resources listed below:

Look into the Revision Notes on Solution of Triangles for a quick revision.

Various Recommended Books of Mathematics are just a click away.

```
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 2,385 off
USE CODE: COUPON10