In right triangle ABC, m∠C = 90°. if cos B = 5/13, which function also equals 5/13? I need help quick please! In right triangle ABC, m∠C = 90°. if cos B = 5/13, which function also equals 5/13? Sin A= 5/13# [Ans] Answer link. Related questions. How do I determine the molecular shape of a molecule? TrigonometricIdentities (1) Conditional trigonometrical identities We have certain trigonometric identities. Like sin2 θ + cos2 θ = 1 and 1 + tan2 θ = sec2 θ etc. Such identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called [] 13 Sin(-a) cos(-a)(tg a+ ctg a) упростить Найдём его площадь: S=a*b=5*2=10(кв.ед). По бокам мы видим два прямоугольных треугольника. Для того, чтобы найти площадь трапеции, нужно отнять от площади Section5.3 Double-Angle, Power-Reducing, and Half-Angle Formulas 611 Solution We will apply the formula for twice. Use with Square the numerator: Square the denominator. 1 4 1 2 1 4 =+cos 2x+ cos2 2x We can reduce the power of . Open in Appwe have the value of and but we don't have the value of and so, first we find the value of and let side opposite to angle hypotenuse where is any positive integer So, by Pythagoras theorem we can find the third side of a triangle taking positive square root as side cannot be negative So, Base we know that side adjacent to angle hypotenuse so, now we have to find the we know that let side adjacent to angle hypotenuse where is any positive integer so, by Pythagoras theorem, we can find the third side of a triangle taking positive square root since, side cannot be negative so, perpendicular we know that Now putting the values, we get Was this answer helpful? 00 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 $\begingroup$I've used the angle sum identity to end up with $\cos A \cos B -\sin A \sin B = \frac{5}{13} = \frac{3}{5}\cos B -\frac{4}{5} \sin B$, but don't know how to proceed from here. Any tips? asked Aug 10, 2017 at 1020 $\endgroup$ 1 $\begingroup$Hint $\cos B=\cosA+B-A$ use the compound angle formula answered Aug 10, 2017 at 1029 David QuinnDavid gold badges19 silver badges48 bronze badges $\endgroup$ $\begingroup$ $$A=\arcsin\dfrac45=\arccos\dfrac35$$ $$A+B=\arccos\dfrac5{13}=\arcsin\dfrac{12}{13}$$ $$B=\arccos\dfrac5{13}-\arccos\dfrac35$$ answered Aug 10, 2017 at 1056 $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged . Byju's AnswerStandard XIIMathematicsComposition of Trigonometric Functions and Inverse Trigonometric FunctionsIf cos a+b=4 ...QuestionOpen in AppSolutiongiven, cosA+B = 4/5, thus tanA+B=3/4. sinA-B=5/13,thus tanA-B=5/12. then tan2A=tanA+B+A-B =tanA+B+tanA-B/1-tanA+BtanA-B =3/4+5/12/1-3/45/12 =56/ Corrections20Similar questionsQ. If sinA=45 and cosB=513, where 0

sin a 4 5 cos b 5 13