4 สูตรฟั งก์ชนตรี โกณมิติของผลบวกและผลต่างของมุมหรื อจํานวนจริง ั sin ( A B) sin A cos B cos A sin B sin ( A B) sin A cos B cos A sin B cos (A B) cos A cos B sin A sin B cos (A B) cos A cos B sin A sin B tan ( A B) tan (A B
1– 10 Soal Aturan Sinus, Aturan Cosinus dan Luas Segitiga dan Jawaban. 1. Deketahui segitiga ABC, dengan panjang AC = 25 cm, sudut A = 60°, dan sudut C = 75° jika sin 75° = 0,9659, tentukan panjang BC dan AB.
piyushTiwari417904.08.2018 Math Secondary School answered • expert verified Cos (a+b) = 4/5 and sin (a-b) = 5/13 then tan2a is Advertisement Expert-verified answer srilukolluru I hope it will help you mark as brainliest answer Find Math textbook solutions? See all Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3
Wehave six trigonometric ratios Sin, Cos, Tan, Cosec, Sec, Cot. sin A = Perpendicular / Hypotenuse. cos A = Base / Hypotenuse. tan A = Perpendicular / Base. cot A = Base / Perpendicular. sec A = Hypotenuse / Base. cosec A = Hypotenuse / Perpendicular. Here, A is that the angle opposite to the perpendicular side.
If sin A = 4 5 and cos B = − 12 13 where A and B lie in first and third quadrant respectively, then cos (A + B) = View Solution. Solve. Guides
mo m What Did Mrs. Margarine Think About Her Sister' Husband? For each exercise, select the correct ratio from the four choices given. Write the letter of the correct choice
Różnezadania z trygonometrii. Szybka nawigacja do zadania numer: 10 20 30 40 50 60 70 . W tym nagraniu wideo omawiam typowe zadanie z trygonometrii, w którym mamy daną wartość jednej funkcji trygonometrycznej, a musimy policzyć wartości wszystkich pozostałych funkcji trygonometrycznych. Zadania tego typu można rozwiązywać na kilka
let sin A = 5/13 with A in Quadrant I and sin B = 15/17 with B in Quadrant I find the following. sin(A + B) Find sin s. cos s = 5 over 13 and s is in quadrant l. Find sin s. cos s = 1 over 2 and s is in quadrant I. Find cos (2A), if sin (A) = 12 over 13, and A is in quadrant 2.
a2 = b 2 + c 2 − 2bc cos(A) b 2 = a 2 + c 2 − 2ac cos(B) c 2 = a 2 + b 2 − 2ab cos(C) But it is easier to remember the "c 2 =" form and change the letters as needed ! As in this example: Example: Find the distance "z" The letters are different! But that doesn't matter. We can easily substitute x for a, y for b and z for c
Ifcos (α + β) = 4/5 and sin (α – β) = 5/13, where 0 ≤ α, β ≤ π/4, then tan 2α is equal to. 1) 25/16. 2) 56/33. 3) 19/12. 4) 20/7. Answer: (2) 56/33. Solution: Given, cos(α + β) = 4/5. ⇒ α + β ∈ Ist quadrant. sin(α – β) = 5/13. ⇒ α − β ∈ Ist quadrant {since 0 ≤
GNJeTB. The correct option is D-1665Explanation for the correct 1 Find the value of cosA,sinBGiven that, sinA=45and cosB= know that, sin2θ+cos2θ=1cosA=1-sin2A=1-452=35Now the value of sinBis negative because B lies in 3rd quadrant. sinB=1-12132=1-144169=25169=-513Step 2 Find the value of cosA+BWe know that, cosA+B= option D is correct.
Given as sin A = 4/5 and cos B = 5/13 As we know that cos A = √1 – sin2 A and sin B = √1 – cos2 B, where 0 < A, B < π/2 Therefore let us find the value of sin A and cos B cos A = √1 – sin2 A = √1 – 4/52 = √1 – 16/25 = √25 – 16/25 = √9/25 = 3/5 sin B = √1 – cos2 B = √1 – 5/132 = √1 – 25/169 = √169 – 25/169 = √144/169 = 12/13 i sin A + B As we know that sin A +B = sin A cos B + cos A sin B Therefore, sin A + B = sin A cos B + cos A sin B = 4/5 × 5/13 + 3/5 × 12/13 = 20/65 + 36/65 = 20 + 36/65 = 56/65 ii cos A + B As we know that cos A +B = cos A cos B – sin A sin B Therefore, cos A + B = cos A cos B – sin A sin B = 3/5 × 5/13 – 4/5 × 12/13 = 15/65 – 48/65 = -33/65 iii sin A – B As we know that sin A – B = sin A cos B – cos A sin B Therefore, sin A – B = sin A cos B – cos A sin B = 4/5 × 5/13 – 3/5 × 12/13 = 20/65 – 36/65 = -16/65 iv cos A – B As we know that cos A - B = cos A cos B + sin A sin B Therefore, cos A - B = cos A cos B + sin A sin B = 3/5 × 5/13 + 4/5 × 12/13 = 15/65 + 48/65 = 63/65
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following sin A + B Given \[ \sin A = \frac{4}{5}\text{ and }\cos B = \frac{5}{13}\]We know that\[ \cos A = \sqrt{1 - \sin^2 A}\text{ and }\sin B = \sqrt{1 - \cos^2 B} ,\text{ where }0 < A , B < \frac{\pi}{2}\]\[ \Rightarrow \cos A = \sqrt{1 - \left \frac{4}{5} \right^2} \text{ and }\sin B = \sqrt{1 - \left \frac{5}{13} \right^2}\]\[ \Rightarrow \cos A = \sqrt{1 - \frac{16}{25}}\text{ and }\sin B = \sqrt{1 - \frac{25}{169}}\]\[ \Rightarrow \cos A = \sqrt{\frac{9}{25}}\text{ and }\sin B = \sqrt{\frac{144}{169}}\]\[ \Rightarrow \cos A = \frac{3}{5}\text{ and }\sin B = \frac{12}{13}\]Now,\[ \sin\left A + B \right = \sin A \cos B + \cos A \sin B\]\[ = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} + \frac{36}{65}\]\[ = \frac{56}{65}\]
