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## Mathematics

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**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).