Mathematics

Properties of Triangle - 2 Session Properties of Triangle - 2

Session Objectives

Solution of Right-angled Triangle Session Objective Solution of Right-angled Triangle Solution of a Oblique Triangle (a) When three sides are given (b) When three angles are given (c) When two sides and the included angle between them are given Two angles and one of the corresponding sides are given (e) When two sides and an angle opposite to one of them is given

Introduction A triangle has three sides and three angles. If three parts of a triangle, at least one of which is side, are given then other three parts can be uniquely determined. Finding other unknown parts, when three parts are known is called ‘solution of triangle’.

Solution of Right-angled Triangle (a) Given a side and an acute angle Let the angle A (acute) and side c of a right angle at C be given c b a C A B or b = c sinB, a = c cosB

Solution of Right-angled Triangle (b) Given two sides Let a and b are the sides of .C is the right angle. Then we can find the remaining angles and sides by the following way c a b C B A

Solution of a Oblique Triangle (a) When three sides are given If the given data is in sine, use the following formulae If the given data is in cosine, use the following formulae.

Solution of a Oblique Triangle If the given data is in tangent, then we use (iv) If the lengths of the sides a, b and c are small, the angle of triangle can also be obtained by cosine rule. (v) For logarithmic computation, we define the following.

Solution of a Oblique Triangle (b) When three angles are given: In this case, the sides cannot be determined uniquely. Only ratio of the sides can be determined by sine rule and hence there will be infinite number of such triangles.

Solution of a Oblique Triangle (c) When two sides and the included angle between them are given: If two sides b and c and the included angle A are given, then (B – C) can be found by using the following formula: If b < c, then we use

Solution of a Oblique Triangle Two angles and one of the corresponding sides are given: The value of the other side and remaining angle can be found by

Solution of a Oblique Triangle (e) When two sides and an angle opposite to one of them is given: In this case, either no triangle or one triangle or two triangles are possible depending on the given parts. Therefore, this case is known as ambiguous case. Let a, b and the angle A are given.

Solution of a Oblique Triangle This is a quadratic equation in c. Let c1 and c2 be two values of c

Solution of a Oblique Triangle Case I: When a < b sinA Hence, from (i) and (ii), become imaginary. No triangle is possible.

Solution of a Oblique Triangle Case II: When a = b sinA  From (i) and (ii), But will be positive when A is acute angle. In this case, only one triangle is possible provided is acute.

Solution of a Oblique Triangle Also a = b sinA The triangle is right-angled in this case. Case III: When a > b sinA  From (i) and (ii), are real.

Solution of a Oblique Triangle But triangle is possible only when are positive. For this we have to consider the following cases: (a) When a > b Hence only one triangle is possible.

Solution of a Oblique Triangle (b) When a = b Hence, only one triangle is possible.

Solution of a Oblique Triangle (c) When a < b  Two triangles are possible. (Ambiguous case) Thus, when a, b and are given, two triangles are possible when a > b sinA and a < b [For ambiguous case]

Class Test

Class Exercise - 1 If the sides of a triangle are Prove that its largest angle is 120°.

Solution  a, b, c are the sides of the triangle,

Solution Cont. Hence, a, b, c are positive, when x > 1Now as x > 1 a – c = x2 + x + 1 – x2 + 1 = x + 2 > 0 as x > 1  a is the largest side  A is the largest angle.

Solution Cont.

Class Exercise - 2 The angles of a triangle are in the ratio 1 : 2 : 7. Show that the ratio of the greatest side to the least sides is

Solution Let the angles of the triangle be x°, 2x°, 7x°.  x + 2x + 7x = 180°  x = 18°  A = 18°, B = 36°, C = 126°  Least side is a and the greatest side is c.

Solution Cont. Proved.

Class Exercise - 3 Let The number of triangle such that log b + 10 = log c + L sinB is one (b) two (c) infinite (d) None of these

Solution  log b + 10 = log c + L sin B  log b + 10 = log c + 10 + log sinB  log b = log (c sinB)  b = c sinB  Only one triangle is possible

Class Exercise - 4 In the ambiguous case, if the remaining angles of the triangles formed with a, b and A be (a) 2 cosA (b) cosA (c) 2 sinA (d) sinA

Solution The two triangles formed are

Solution Cont. Triangles are formed with a, b and A,

Solution Cont. Here two values of c are . Hence answer is (a).

Class Exercise - 5 In ambiguous case, where b, c, B are given and prove that

Solution b, c and B are given, This is quadratic equation in a.

Solution Cont. It is given that

Solution Cont. (Negative sign is neglected as ‘b’ is the length of side of triangle).

Thank you