Session Properties of Triangle - 2
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’.
B c a A C b 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 or b = c sinB, a = c cosB
Let a and b are the sides of .C is theright angle. Then we can find the remaining angles and sides by the following way A c b B C a Solution of Right-angled Triangle (b) Given two sides
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
Hence, from (i) and (ii), become imaginary. Solution of a Oblique Triangle Case I: When a < b sinA No triangle is possible.
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 Case II: When a = b sinA From (i) and (ii),
From (i) and (ii), are real. Solution of a Oblique Triangle Also a = b sinA The triangle is right-angled in this case. Case III: When a > b sinA
But triangle is possible only when are positive. For this we have to consider the following cases: Solution of a Oblique Triangle (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.
Thus, when a, b and are given, two triangles are possible when a > b sinA and a < b [For ambiguous case] Solution of a Oblique Triangle (c) When a < b Two triangles are possible. (Ambiguous case)
If the sides of a triangle are Prove that its largest angle is 120°. Class Exercise - 1
Solution a, b, c are the sides of the triangle, a > 0, b > 0, c > 0
as x > 1 • a – c = x2 + x + 1 – x2 + 1 = x + 2 > 0 as x > 1 Solution Cont. Hence, a, b, c are positive, when x > 1Now a is the largest side Ais the largest angle.
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 Class Exercise - 2
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.
Let The number of triangle such that log b + 10 = log c + L sinB is • one (b) two • (c) infinite (d) None of these Class Exercise - 3
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
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 Class Exercise - 4
The two triangles formed are Solution
Triangles are formed with a, b and A, Solution Cont.
Here two values of c are . Solution Cont. Hence answer is (a).
In ambiguous case, where b, c, B are given and prove that Class Exercise - 5
b, c and B are given, Solution This is quadratic equation in a.
It is given that Solution Cont.
Solution Cont. (Negative sign is neglected as ‘b’ is the length of side of triangle).