🐚 Sin A 4 5 Cos B 5 13

Question888014: Find the exact value of sin(a-B) if sin a=(5/13) and sin B=(8/17) and a and B are between 0 and pi/2 Answer by stanbon(75887) (
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,\[ \sin\left A + B \right = \sin A \cos B + \cos A \sin B\]\[ = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} + \frac{36}{65}\]\[ = \frac{56}{65}\] 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. $\begingroup$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 .
Click here:point_up_2:to get an answer to your question :writing_hand:if sin a frac45 and cos b frac1213 where both a and b. Solve. Guides. Join / Login. Use app Login. Standard XII. Mathematics. Local Maxima. Question. If sin A = 4 5 and cos B = 12 13, Where both A and B lie in first quadrant. If sin A = 4 5 and cos B = - 12 13,
We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169⇒cosA=35 and sinB=1213 Now, sinA+B=sinA cosB+cosA sinB =45×513+35×1213=2065+3665=20+3665=5665 ii We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1452 and sin B=√1−5132 ⇒cosA=√1−1615 sin B=√1−25169 ⇒cosA=√25−1625 and sin B=√169−25169 ⇒cosA=√925 and sin B=√144169 cosA=35 and sinB=1213 Now, cosA+B=cosA cosB−sinA sinB =35×513−45×1213 =1565−4865 =15−4865=−3365 iii We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169 ⇒cosA=35 and sinB=1213 Now, sinA−B=sinA cosB−cosA sinB =45×513−35×1213 =2065+3665=20−3665=−1665 iv We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=169−25169 ⇒cosA=√925 and sinB=1213 Now, cosA−B=cosA cosB+sinA sinB =35×513+45×1213 =1565+4865 =15+4865 =6365
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Answer:cos (A+B)=16/65. Step-by-step explanation: Using Pythagoras theory and the idea of "SOH,CAH, TOA "cosA=3/5 cosB=12/13. Cos(A+B)=cosAcosB-singing. Cos (A+B)=(3/5)(12/13)-(4/5)(5/13) =36/65-20/65 =16/65 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.

1comment for "Diketahui sin A = 4/5 dan cos B = 1/3√2 dengan A dan B sudut lancip. Tentukan nilai:" K October 7, 2021 at 9:36 AM Delete Comment diketahui sin a = 1/2 dan cos b = 1/2 akar 2 dengan a sudut tumpul dan b sudut lancip. tentukan tan (a + b) ! Newer Posts Older Posts Pondok Budaya Bumi Wangi. DMCA. About Me

The correct option is B5633Explanation for the correct optionStep 1. Find the value of tan2αGiven, cosα+β=45⇒ sinα+β=35 sinα-β=513⇒ cosα-β=1213Now, we can write2α=α+β+α–βStep 2. Take "tan" on both sides, we gettan2α=tanα+β+α–βtan2α=[tanα+β+tanα–β][1–tanα+βtanα–β] …1 ∵tanθ+ϕ=tanθ+tanϕ1-tanθtanϕAlso,tanα+β=sinα+βcosα+β=3/54/5=34tanα–β=sinα–βcosα–β=5/1312/13=512Step 3. Put these values in equation 1, we get∴tan2α=3/4+5/121–3/45/12=9+5/1248–15/48=5633Hence, Option ‘B’ is Correct.
Introductionto Trigonometry Class 10 Important Questions Very Short Answer (1 Mark) Question 1. If tan θ + cot θ = 5, find the value of tan2θ + cotθ. (2012) Solution: tan θ + cot θ = 5 . [Given. tan 2 θ + cot 2 θ + 2 tan θ cot θ = 25 . [Squaring both sides. tan 2
>>Class 11>>Maths>>Trigonometric Functions>>Trigonometric Functions of Sum and Difference of Two angles>>If cos A = 4/5 , cos B = 12/13 , 3pi/Open in AppUpdated on 2022-09-05SolutionVerified by TopprA and B both lie in the IV quadrant.=> are negativei iiSolve any question of Trigonometric Functions with-Was this answer helpful? 00More From ChapterLearn with Videos Practice more questions
Ifα and β lie between 0 and (π/4), find tan 2α, given that cos(α+β)=(4/5) and sin(α-β)=(5/13). (A) 50/33 (B) 56/33 (C) 52/33 (D) 48/33. Check

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

sinA = 4/5 cos A = 3/5 sin B =12/13 cos B = 5/13 Periksa ulang, Sudut A tumpul sehingga berada di kuadran II (antara 90 dan 180) . Lihat ilustrasi di bawah, untuk kuadran II nilai sin adalah positif, sehingga sin A benar 4/5. Sementara untuk cos A, karena dikuadran II, nilainya negatif, jadi cos A = − 3/5
/ Fórmulas / Matemática / 1. Relações trigonométricas fundamentais $\mathrm{sen}^{2} a + \cos^{2} a = 1$ $tg a = \frac{sen a}{\cos a}$ $cotg a = \frac{\cos a }{sen a}$ $sec a = \frac{1}{\cos a}$ $cossec a = \frac{1}{sen a}$ 2. Relações trigonométricas derivadas $tg^{2} a + 1 = sec^{2} a$ $cotg^{2} a +1 = cossec^{2} a$ 3. Seno da soma - Cosseno da soma - Tangente da soma $sena+b = sena \ . \cos b + senb \ . \cosa$ $\cos a+b = \cos a \ . \cos b - sena \ . senb$ $tga+b = \frac{tga + tgb}{1-tga \ . tgb}$ 4. Seno da diferença - Cosseno da diferença - Tangente da diferença $sena-b = sena \ . \cos b - senb \ . \cos a$ $\cos a-b = \cos a \ . \cos b + sena \ . senb$ $tga-b = \frac{tga - tgb}{1+tga \ . tgb}$ 5. Soma de senos - Soma de cossenos - Soma de tangentes $sen a + sen b = 2 sen \left \frac{a+b}{2} \right \ . \cos \left \frac{a-b}{2} \right$ $ \cos a+ \cos b = 2 \cos \left\frac{a+b}{2} \right \ . \cos \left\frac{a-b}{2}\right$ $tg a + tg b = \left \frac{sen a+b}{\cos a \ . \cos b} \right$ 6. Subtração de senos - Subtração de cossenos - Subtração de tangentes $ sen a - sen b = 2 sen \left \frac{a-b}{2} \right \ . \cos \left \frac{a+b}{2} \right $ $ \cos a - \cos b = -2 sen \left \frac{a+b}{2} \right \ . sen \left \frac{a-b}{2} \right$ $tg a -tg b = \left \frac{sen a-b}{\cos a \ . \cos b} \right $ 7. Arco metade $sen \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1- \cos a}{2}}$ $\cos \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1+\cos a}{2}}$ $tg \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1- \cos a}{1+ \cos a}}$ 8. Arco duplo $sen2a = 2sena \ . \cos a$ $\cos 2a = \cos^{2} a - sen^{2}a$ $tg2a = \frac{2tga}{1-tg^{\style{font-familyArial; font-size31px;}{2}}a}$ 9. Arco triplo $sen3a = 3sena-4sen^{3}a$ $\cos 3a = 4 \cos^{3} 3a - 3 \cos a$ $tg 3a = \frac{3tg a-tg^{3}a}{1-3tg^{\style{font-familyArial; font-size30px;}2}a}$ 10. Arco quádruplo $sen4a =4sena \ . \cos a -8sen^{3} a \ . \cos a $ $\cos 4a = 8 \cos^{4} a - 8 \cos^{2} a +1$ $tg 4a = \frac{4tg a- 4tg^{3}a}{1-6tg^{\style{font-familyArial; font-size30px;}2}a+tg^{\style{font-familyArial; font-size30px;}4} a}$ 11. Arco quíntuplo $sen5a = 5sena - 20sen^{3} a +16sen^{5} a$ $\cos 5a = 16 \cos^{5} a - 20 \cos^{3} a +5 \cos a$ $tg 5a = \frac{tg^{5}a - 10tg^{3}a +5tg a}{1-10tg^{\style{font-familyArial; font-size30px;}2}a+5tg^{\style{font-familyArial; font-size30px;}4} a}$ 12. Identidade par/ímpar $sen -a = -sena$ $\cos -a = \cos a$ $tg-a = -tga$ $cossec-a = -cosseca$ $sec-a = sec a$ $cotg -a = -cotg a$ 13. Arcos complementares $sen 90° \hspace{ -a = \cos a$ $\cos 90° \hspace{ -a = sen a$ $tg 90° \hspace{ -a = cotg a$ $cotg 90° \hspace{ -a = tg a$ $sec 90° \hspace{ -a = cossec a$ $cossec 90° \hspace{ -a = sec a$ 14. Periodicidade $sen 360° \hspace{ +a = sen a$ $\cos 360° \hspace{ +a = \cos a$ $tg 180° \hspace{ +a = tga$ $cotg 180° \hspace{ +a = cotga$ $sec 360° \hspace{ +a = seca$ $cossec 360° \hspace{ +a = cosseca$ 15. Transformação de produto para soma $sen a \ . sen b = \frac { \cos a-b - \cosa+b}{2}$ $\cos a \ . \cos b = \frac {\cos a-b + \cos a+b}{2}$ $sen a \ . \cos b = \frac {sen a-b+sen a+b}{2}$ $tg a \ . tgb = \frac {tg a + tgb}{cotga + cotgb}$ $cotga \ . cotgb = \frac {cotga + cotgb}{tg a + tg b}$ $tga \ . cotgb = \frac {tg a + cotg b}{cotg a + tg b}$ 16. Potências de seno e cosseno $sen^{2} a = \frac{1-cos 2a}{2}$ $sen^{3} a = \frac{3sen a -sen3a}{4}$ $sen^{4} a = \frac{\cos 4a -4 \cos 2a + 3}{8}$ $sen^{5} a = \frac{10sen a -5 sen 3a + sen5a}{16}$ $sen^{6} a = \frac{10 - 15 \cos 2a +6 \cos 4a -cos 6a}{32}$ $\cos^{2} a = \frac{1+ \cos 2a}{2}$ $\cos^{3} a = \frac{3 \cos a +cos3a}{4}$ $\cos^{4} a = \frac{\cos 4a +4 \cos 2a + 3}{8}$ $\cos^{5} a = \frac{10 \cos a +5 sen 3a + \cos 5a}{16}$ $\cos^{6} a = \frac{10 + 15 \cos 2a +6 \cos 4a + cos 6a}{32}$ 
In a right triangle, one angle measures x ∘ ‍ , where cos ⁡ x ∘ = 5 13 So, cos(60) = sin(30), cos(45) = sin(45), and so on. For the other quadrants, you can draw a little picture, and using your knowledge that sine is the vertical distance from the point on the unit circle to the origin and cosine is the horizontal, you can fill in given, cosA+B=4/5, thus tanA+B=3/4 from triangle 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 = 56/33.
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Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β. 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
诱导公式口诀奇变偶不变，符号看象限”意义：. k×π/2±a (k∈z)的三角函数值. （1）当k为偶数时，等于α的同名三角函数值，前面加上一个把α看作锐角时原三角函数值的符号；. （2）当k为奇数时，等于α的异名三角函数值，前面加上一个把α看作锐角时原三角
a sin 15º b. cos 75º c. tan 285º; Diketahui sin A = 12/13 dan cos B = 3/5 dengan sudut A tumpul dan sudut B lancip tentukan a. sin ( A + B ) b. cos ( A + B ) c. tan ( A + B ) Bentuk sederhana dari a. cos 70 cos 20 – sin 70 sin 20 b. sin 160 cos 10 – cos 160 sin 10; Diketahui cos ( A + B ) = ½ dan sin A sin B = ¼ tentukan nilai dari Trigonometry. Trigonometry is a branch of mathematics concerned with relationships between angles and ratios of lengths. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. The Greeks focused on the calculation of chords, while mathematicians in India created the earliest
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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?

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