Questions and Answers #2 Non-Right Triangles and Trigonometry

Questions and Answers Sheet 2 Non-Right Triangles and Trigonometry Question #1 Determine whether the two triangles are similar. Explain 1. The two triangles are not similar. Since 𝑚∠𝐵 ≠ 𝑚∠𝐹 , the triangles are not similar by the AA criterion. 2. The two triangles are similar. By the Interior angles theorem, 𝑚∠𝐶 = 67∘ , so the triangles are similar by the AA criterion. 3. The two triangles are similar. By the interior angles theorem, 𝑚∠𝑐 = 57∘ , so the triangles are similar by the AA similarity criterion. 3. The two triangles are not similar. By the interior angles theorem, 𝑚∠𝐶 = 57∘ so, the triangles are not similar by the AA similarity criterion. Answer: Step 1 In △ 𝐴𝐵𝐶 ∠𝐵 + ∠𝐴 + ∠𝐶 = 180∘ 48∘ + 65∘ + ∠𝐶 = 180∘ ∠𝐶 = 180∘ − (48 + 65∘ ) ∠𝐶 = 67∘ Step 2 In △ 𝐷𝐸𝐹 ∠𝐷 + ∠𝐸 + ∠𝐹 = 180∘ 65∘ + ∠𝐸 + 67∘ = 180∘ ∠𝐸 = 180∘ − (67∘ + 65∘ ) ∠𝐸 = 48∘ Now in △ 𝐷𝐸𝐹 and △ 𝐴𝐵𝐶 ∠𝐷 = ∠𝐴 = 65∘ ∠𝐸 = ∠𝐵 = 48∘ ⇒△ 𝐷𝐸𝐹 ∼△ 𝐴𝐵𝐶 [By AA criteria] Question #2 Prove that if a line is drawn parallel to the base of a triangle, it cuts off a triangle similar to the given triangle. Given: △ 𝐴𝐵𝐶, 𝐷𝐸|| 𝐴𝐵 Prove: △ 𝐷𝐸𝐶 ∼△ 𝐴𝐵𝐶 Statements Reasons 1. (see above) 1. Given 2. 2. If two parellel lines are cut by a transversal, then the corresponding angles are equal. 3. DEC ~ ABC 3. Answer: Step 1 Prove that, if a line is drawn parallel to the base of a triangle, it cuts off a triangle similar to the given triangle as follows. Step 2 Statements 1. ABC , DE || AB 2.CDE = CAB, CED = CBA and ACB = DCE 3. DEC ~ ABC Reasons 1. Given 2. If two parellel lines are cut by a transversal, then the corresponding angles are equal. 3. If two triangles have the three angles of one equal respectively to the three angles of the other, then the triangles are similar. Question #3 Prove that two isosceles triangles are similar if any angle of one equals the corresponding angle of the other. a. Case 1: Equal Vertex Angles Given Isosceles △ 𝐴𝐵𝐶 , isosceles △ 𝐷𝐸𝐹, 𝐴𝐶 = 𝐵𝐶, 𝐷𝐹 = 𝐸𝐹, ∠𝐶 = ∠𝐹 Prove: △ 𝐴𝐵𝐶 ∼△ 𝐷𝐸𝐹 Statements 1.( see above) 2. 3. ABC ~ DEF Answer: Step 1 Reasons 1. Given 2. Isosceles triangles with equal vertex angles have equal base angles also. 3. Consider the figure: ∠𝐴𝐶𝐵 = ∠𝐷𝐹𝐸 AC=BC DF=EF Step 2 Find the statement for isosceles triangles with equal vertex angles have equal base angles. Given: ∠𝐴𝐶𝐵 = ∠𝐷𝐹𝐸 AC=BC DF=EF If two isosceles triangles with equal vertex angles, then they have equal base angles also. Since, ∠𝐴𝐶𝐵 = ∠𝐷𝐹𝐸 ∠𝐶𝐴𝐵 = ∠𝐹𝐷𝐸 and, ∠𝐶𝐵𝐴 = ∠𝐹𝐸𝐷 2.CAB = FDE CBA = FED 2. Isosceles triangles with equal vertex angles have equal base angles also. Step 3 Find the reason of similarity of given triangles. Use theorem 57, If two triangles have the three angles of one equal respectively to the three angles of the other, then the triangles are similar. Corollary 57-1 If two angles of one triangle are equal respectively to two angles of another, then the triangles are similar. (a.a) Theorem 58-If two triangles have two pairs of sides proportional and the included angles equal respectively, then the two triangles are similar. (s.a.s) ∠𝐴𝐶𝐵 = ∠𝐷𝐹𝐸 (Theorem 57) ∠𝐶𝐴𝐵 = ∠𝐹𝐷𝐸 and ∠𝐶𝐵𝐴 = ∠𝐹𝐸𝐷 (Corollary 57-1) 𝐴𝐶 𝐷𝐹 = So, 𝐵𝐶 𝐸𝐹 and All above equal angles (Theorem 58) △ 𝐴𝐵𝐶 =△ 𝐷𝐸𝐹 3. ABC = DEF 3. If two triangles have three angles of one equal respectively to the three angles of the other, then the triangles are similar. Question #4 Triangles ABC and DEF are right triangles, as shown △ 𝐴𝐵𝐶 is similar to △ 𝐷𝐸𝐹 Which ratios are equal to cos(𝐵) ? Choose the TWO ratios that apply. Answer: Solution: Given that △ 𝐴𝐵𝐶 and △ 𝐷𝐸𝐹 are right triangle and △ 𝐴𝐵𝐶 ≈△ 𝐷𝐸𝐹 Now as we know 𝐵𝑎𝑠𝑒 cos 𝜃 = cos 𝐵 = 𝐻𝑦𝑝𝑜𝑡𝑎𝑛𝑐𝑜𝑢𝑠 𝐵𝐶 𝐴𝐵 Now by similarity ratio property we know 𝐴𝐵 𝐵𝐶 = 𝐷𝐸 ⇒ 𝐸𝐹 𝐷𝐸 𝐸𝐹 = 𝐵𝐶 𝐴𝐵 from (1) and (2) 𝐸𝐹 cos 𝐵 = 𝐷𝐸 Question #5 The diagonals of a trapezoid divide each other proportionally. Given: Trapezoid ABCD with diagonal AC, BD intersecting at O Prove: 𝐴𝑂 𝐶𝑂 = 𝐵𝑂 𝐷𝑂 Statements Reasons 1.( see above) 1. Given 2. 2. 3.(Give 2 pairs of equal angles) 3. If two parallel lines are cut by a transveral, then the alternate interior angles are=. 4. AOB ~ COD 4. AO BO 5. = 5. CO DO Answer: Statements Reasons 1.Diagonal AC, BD intersecting O 1. Given 2. AB || CD 2.ABCD is Trapezoid 3.BDC = DBA and DCA = CAB 3. If two parallel lines are cut by a transveral, then the alternate interior angles are=. 4. AOB ~ COD 4.If two angles of one triangle are equal respectively to two angles of another, then the triangles are similar. (A.A) AO BO 5. = 5.C.S.S.T.P-corresponding sides of CO DO similar triangles are proportional Question #6 Given b=2, a=3 , and B=40 (degrees), determine whether this information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangles. Answer: Step 1 Given that, the values are b=2, a=3 and 𝐵 = 40∘ Recall that, the sine formula is sin 𝐴 𝑎 = 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝑐 sin 𝐵 𝑏 sin 𝐴 = 𝑎 ( 𝐴 = sin −1 sin 𝐵 ( ) 𝑏 𝑎⋅sin 𝐵 𝑏 ) Step 2 Substitute a=3, b=2 and 𝐵 = 40∘ in 𝐴 = sin−1 ( 𝐴 = sin−1 ( = sin 3⋅sin(40∘ ) 2 −1 (0.9642) = 1.3023 × ) 180∘ 𝜋 = 74.6162∘ 𝐴 = 74.6∘ (or) 𝐴 = 180∘ − 74.6∘ 𝐴 = 105.4∘ (∵ 105.4∘ + 40∘ < 180∘ ) Step 3 Let 𝐴 = 74.6∘ then the angle of C is, 𝐶 = 180∘ − 74.6∘ − 40∘ = 65.4∘ Use the sine formula to find the value of c. sin 𝐵 𝑏 𝑐= = sin 𝐶 𝑐 𝑏 sin 𝐶 sin 𝐴 𝑎⋅sin 𝐵 𝑏 ) 2 sin 65.4∘ 𝑐= sin 40∘ c=2.83 Thus, the possible values of the triangle is a=3, b=2, c=2.83 and 𝐴 = 74.6∘ , 𝐵 = 40∘ , 𝐶 = 65.4∘ Step 4 Let 𝐴 = 105.4∘ then the angle of C is, 𝐶 = 180∘ − 105.4∘ − 40∘ = 34.6∘ Use the sine formula to find the value of c. sin 𝐵 𝑏 𝑐= 𝑐= = sin 𝐶 𝑐 𝑏 sin 𝐶 sin 𝐴 2 sin 34.6∘ sin 40∘ 𝑐 = 1.77 Thus, the possible values of the triangle is a=3, b=2, c=1.77 and 𝐴 = 105.4∘ , 𝐵 = 40∘ , 𝐶 = 34.6∘ Therefore, the given values are possible for two triangles. Question #7 Prove that |𝑎 − 𝑏| ≥ |𝑎| − |𝑏|. Answer: It is called the reverse triangle inequality. |𝑎 − 𝑏| ≥ ||𝑎| − |𝑏|| |𝑎| = |𝑎 − 𝑏 + 𝑏| ≤ |𝑎 − 𝑏| + |𝑏| |𝑏| = |𝑏 − 𝑎 + 𝑎| ≤ |𝑎 − 𝑏| + |𝑎| Thus, we have: −|𝑎 − 𝑏| ≤ |𝑎| − |𝑏| ≤ |𝑎 − 𝑏| Question #8 State the number of possible triangles that can be formed using the given measurements 𝑚 < 𝐶 = 62∘ , 𝑏 = 9, 𝑐 = 12 Answer: Step 1 In trigonometry, the formula for the law of sines, or sine rule is the relation between the lengths of the sides of a triangle to the sines of its angles, sin 𝐴 𝑎 sin 𝐵 = 𝑏 = sin 𝐶 𝑐 Step 2 Use the law of sines, to find the value of angle B, substitute angle C with 62∘ , side b with 9 and side c with 12. Multiply both sides by 9 to isolate sin B and further simplify to get the two values of 41.45∘ and 138.55∘ . Then, to find angle A, subtract the sum of angle B and angle C from 180∘ . For ∠ 𝐵 = 41.45∘ , the value of angle A is 76.55∘ and for ∠𝐵 = 138.55∘ , the value of angle A is −20.55∘ and since ∠A cannot be negative. Hence, there is only one triangle possible. sin 𝐵 𝑏 sin 𝐵 9 = = sin 𝐶 𝑐 sin 62∘ sin 𝐵 = =0.662 12 3(0.8829) 4 𝐵 = sin−1 (0.662) = 41.45∘ or 138.55∘ ∠𝐴 = 180∘ − (∠𝐵 + ∠𝐶) = 180∘ − (41.45∘ + 62∘ ) = 180∘ − 103.45∘ = 76.55∘ ∠𝐴 = 180∘ − (∠𝐵 + ∠𝐶) = 180∘ − (138.55∘ + 62∘ ) = 180∘ − 200.55∘ = −20.55∘ Question #9 The perpendicular bisector 𝐴𝐵 in the right triangle △ 𝐴𝐵𝐶 forms the triangle with the area 3 times smaller than the area of △ 𝐴𝐵𝐶. Find the measures of acute angles in △ 𝐴𝐵𝐶 Answer: Step 1 The Sides of AB and AC have unit length Hence, △ 𝐴𝐵𝐶 is isosceles with Congruent angles ∠𝐵 and ∠𝐶 Since 𝑚(∠𝐴) is given as 60 this means that, 𝑚(∠𝐵) + 𝑚(∠𝐶) = 120 and so, 𝑚(∠𝐵) + 𝑚(∠𝐶) = 60 Therefore △ 𝐴𝐵𝐶 is equilateral Equilateral triangles have three lines of reflective symmetry in which the lines joining each vertex to the midpoint of the opposite side. This means that CD is a line of symmetry for △ 𝐴𝐵𝐶 and so CD is perpendicular to AB. Step 2 By applying Pythagorean Theorem to the right triangle △ 𝐴𝐷𝐶 |𝐴𝐷|2 + |𝐶𝐷|2 = |𝐴𝐶|2 We Know that D is the midpoint of AB. B is on the unit circle, Hence |AB|=1 and |AD|=12 Since C is unit circle, We have |AC|=1 Plugging the values for |AD| and |AC| into the formula gives, |𝐶𝐷| = 3√2. Since CD is perpendicular to AB . Therefore, 𝐶(12, 3√2) sin 6 0∘ = 3√2 and cos 6 0∘ = 12 Question #10 Given b=2, a=3, and B=40 (degrees), determine whether this information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangles. Answer: Step 1 Given that, the values are b=2, a=3 and 𝐵 = 40∘ Recall that, the sine formula is sin 𝐴 sin 𝐵 = 𝑎 𝑏 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝑐 sin 𝐴 = 𝑎 ( sin 𝐵 𝐴 = sin−1 ( ) 𝑏 𝑎×sin 𝐵 𝑏 ) Step 2 Substitute a=3, b=2 and 𝐵 = 40∘ in 𝐴 = sin−1 ( 𝐴 = sin−1 ( = sin 3×sin(40∘ ) 2 −1 (0.9642) = 1.3023 × 𝑎×sin 𝐵 𝑏 ) ) 180∘ 𝜋 = 74.6162∘ 𝐴 = 74.6∘ or 𝐴 = 180∘ − 74.6∘ 𝐴 = 105.4∘ (∵ 105.4∘ + 40∘ < 180∘ ) Step 3 Let 𝐴 = 74.6∘ then the angle of C is, 𝐶 = 180∘ − 74.6∘ − 40∘ Use the sine formula to find the value of c. sin 𝐵 𝑏 𝑐= 𝑐= = sin 𝐶 𝑐 𝑏 sin 𝐶 sin 𝐴 2 sin 65.4∘ sin 40∘ c=2.83 Thus, the possible values of the triangle is a=3, b=2, c=2.83 and 𝐴 = 74.6∘ , 𝐵 = 40∘ , 𝐶 = 65.4∘ Step 4 Let 𝐴 = 105.4∘ then the angle of C is, 𝐶 = 180∘ − 105.4∘ − 40∘ = 34.6∘ Use the sine formula to find the value of c. sin 𝐵 𝑏 𝑐= 𝑐= = sin 𝐶 𝑐 𝑏 sin 𝐶 sin 𝐴 2 sin 34.6∘ sin 40∘ c=1,77 Thus, the possible values of the triangle is a=3, b=2, c=1.77 and 𝐴 = 105.4∘ , 𝐵 = 40∘ , 𝐶 = 34.6∘ Therefore, the given values are possible for the triangles.