applying the law of cosines solve the following problems
1. applying the law of cosines solve the following problems
Answer:
sana po makatulong
STAY SAFE
2. Applying the law of cosines solve the following problems
Answer:
Asan po ang mga questions?
Answer:
Where is the answers po?
3. applying the law of cosines, solve the following problems. Round
Answer:
asaan yung sasagutin
Step-by-step explanation:
4. a. To solve the problem using the law of sine, what is needed? __________________b. Show your solution.c. What are the measure of its sides and its perimeter? _______________________d. To solve the problem using the law of cosine what is needed? _________________________________
Answer:
Sa Picture Yung Sagot
Step-by-step explanation:
5. a) Determine if the law of cosines can be applied (be able to justify).b) Formulate an equation in x for items where the law of cosines is applicable.
Answer:
a) The law of Cosines can be applied by using the Case IV :Three sides given.
b) X=133.5°
6. how will you use the law of cosines and the law of sines and solving problems involving oblique triangles
Answer:The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.=============================[tex]hope \: it \: helps.[/tex]
#CarryOnLearning
7. ENGAGEMENT:Applying the law of cosines, solve the following problems. Round to the nearest hundreth.please pakisagot naman po, thankyouuuu!!^^
Answer:
hello poh 1-2 only. that's the answer
number 3 sobrang liit ng letra...
hope it's help po
8. halimbawa ng cosine law problem?
May isang bata lumakad sa kanyang kanan ng 10 metro, may isa pang bata naglakad ng 30 degrees north east. anu ang distansya ng dalawang bata nang nakaratin na sila sa kanilang puwesto. Hope this helps =)
9. pa solve po using law of cosines
Answer:
c= 16
Angle A and B=30°
Step-by-step explanation:
PABRAINLIEST PO10. What are the sine and cosine laws in solving a triangle?
Answer:
The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
11. Solving Problems Involving Oblique Triangles Using The LAW OF COSINESSolve for the missing parts of the following triangles.Angle B = ?Angle C = ? a = ?pls help mee
Solution:
The formula for law of cosines is
a = √(b² + c² - 2bc cos A)
where:
a = side opposite to angle A
b and c = sides adjacent to angle A
A = angle opposite to side a
Listing the given values
b = 9
c = 3
A = 108°
Solving for a
a = √(b² + c² - 2bc cos A)
a = √[9² + 3² - 2(9)(3)(cos 108°)]
a = √[81 + 9 - (-16.6869)]
a = √(81 + 9 + 16.6869)
a = √106.6869
a = 10.3289
a = 10.33
Solving for B
B = Arccos [(a² + c² - b²) / (2ac)]
B = Arccos [10.3289² + 3² - 9² / (2 × 10.3289 × 3)]
B = Arccos [(106.6867 + 9 - 81) / 61.9734
B = Arccos (34.6867 / 61.9734)
B = Arccos (0.5597)
B = 55.96°
Solving for C
C = 180° - (A + B)
C = 180° - (108° + 55.96°)
C = 180° - 163.96°
C = 16.04°
Therefore, the answers are
a = 10.33
B = 55.96°
C = 16.04°
#MathIsFun
12. PERFORMANCE TASKDirection. Write a brief essay using the following question.Are the laws of sines and cosines useful in solving problems that mayoccur in our daily routine?
Answer:
yes , because it helps us to solve our daily problem in money on how to budget it or whatever
Many real-world applications involve oblique triangles, where the Sine and Cosine Laws can be used to find certain measurements. ... The Cosine Law is used to find a side, given an angle between the other two sides, or to find an angle given all three sides. For all other questions, the Sine Law can be used.
13. How will you apply the law of sines and cosines in solving real life problems?
Answer:
Many real-world applications involve oblique triangles, where the Sine and Cosine Laws can be used to find certain measurements. It is important to identify which tool is appropriate. The Cosine Law is used to find a side, given an angle between the other two sides, or to find an angle given all three sides.
Hope it helps you ;)
14. A triangular lot has side lengths of 12 m, 14 m, and 20 m. Which law will you apply to solve the problem?A. law of cosineB. law of sineC. law of secantD. law of tangent
Answer:
D. Law of tangent
Step-by-step explanation:
HOPE IT HELPSMARK AS BRAINLIEST
Answer:
A triangular lot has side lengths of 12 m, 14 m, and 20 m. Which law will you apply to solve the problem?
A. law of cosine
B. law of sine
C. law of secant
D. law of tangent
ANSWER!:
D. law of tangent
I hope that helped
Step-by-step explanation:
15. Can you site real life application of law of cosines? Describe how you can apply the law of cosines in that situation.
Answer:
Use the Law of Sines to model and solve real-life problems. You can use the Law of Sines to solve real-life problems involving oblique triangles. For instance, in Exercise 44 on page 438, you can use the Law of Sines to determine the length of the shadow of the Leaning Tower of Pisa.
Step-by-step explanation:
16. Your task is to solve the length of the other side of the lot to allocate an area for the flower landscape. Solve the problem applying the Laws of Cosines.
Answer:
sana po makatulong
STAY SAFE
17. Applying the law of cosines, solve the following problems. Round to the nearest hundredth. 2. Solve ∆ABC, if C = 28°, a = 42, and b = 57.
Answer:
a = 15.98
B = 24.2°, C = 43.8°
________________________________Solution po nasa pic
[tex] \sf{Hope \: that \: helps \: you:)}[/tex]
18. Create a problem that illustrates the law of sine and cosines and solve it please
happy to help you
hope this help
19. applying the law of cosines. solve the following problems round to the nearest hundredth
Answer:
1. Solve ∆ABC.
2.Solve ∆ABC, if C = 28°,a = 42, and b = 57.
3.Solve ∆ABC.
4.Solve ∆RST, given r = 29, s = 16, and t = 24.
pa brainliest po
20. E. Engagement (Time Frame: 40 minutes ) Applying the law of cosines, solve the following problems. Round to the nearest hundredth. 1. Solve ΔABC. 2. Solve ΔABC, if C = 28°, a = 42, and b = 57. 3. Solve ΔABC. 4. Solve ΔRST, given r = 29, s = 16, and t = 24.
[tex]\large\color{black}{{\underline{\bold{Answer} } } }[/tex]
[tex]______________________________________[/tex]
[tex]\small\color{black}{{\underline{\bold{1.){a}^{2} \sqrt{ {b}^{2} + {c}^{2} - 2(b)(c) \cos(A) } } } }} \\ = {a}^{2} \sqrt{ {7}^{2} + {12}^{2} - 2(7)(12) \cos(112)}\\=\small\color{black}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:a=15.99\:\:\:\: }}}} [/tex]
[tex] \small\color{black}{{\underline{\bold{<B \sin {}^{ - 1} = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} } } } } \\ = \sin( {}^{ - 1} ) = \frac{15.99 {}^{2} + {7}^{2} - {12}^{2} }{ {2(15.99)(7)} } \\ = \small\color{black}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:\<B=44.13\:\:\:\: }}}}[/tex]
[tex]\small\color{black}{{\underline{\bold{2.) {c}^{2} \sqrt{ {42}^{2} + {57}^{2} - 2 \sqrt{(42)(57) \cos(28) } } } } }} \\ = \small\color{black}{{{\boxed{\tt\red{} \:\:\:\:\:c=28.03\:\:\:\: }}}}[/tex]
[tex]\small\color{black}{{\underline{\bold{ {\cos { }^{-1} = \frac{ {42}^{2} + {57}^{2} - {28.03}^{2} }{2(42)(57)} } }}} }\\=\small\color{black}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:\<B=28.01\:\:\:\: }}}}[/tex]
[tex]______________________________________[/tex]
21. the law of cosines allows you to solve a triangle as long as you know either of the following
answer; this my anwer thx
22. Explain the step by step procedure in solving problems involving laws of cosine.
Well, it helps to know it's the Pythagoras Theorem with something extra so it works for all triangles:
Pythagoras Theorem:
(only for Right-Angled Triangles) a2 + b2 = c2Law of Cosines:(for all triangles) a2 + b2 − 2ab cos(C) = c2So, to remember it:
think "abc": a2 + b2 = c2,then a 2nd "abc": 2ab cos(C),and put them together:
a2 + b2 − 2ab cos(C) = c2When to UseThe Law of Cosines is useful for finding:
the third side of a triangle when we know two sides and the angle between them (like the example above)the angles of a triangle when we know all three sides (as in the following example).
Answer
The Law of Cosines (also called the Cosine Rule ) says: ... Now we use our algebra skills to rearrange and solve: ... It took quite a few steps, so it is easier to use the "direct" formula (which is just a ...
Carry on learning23. Solving Problems Involving Oblique Triangles Using The LAW OF COSINESSolve for the missing parts of the following triangles.Angle D = ?Angle P = ?I = ?pls help mie
Answer:
see attached pictures
Step-by-step explanation:
24. Are the laws of sines and cosines useful in solving problems that may occur in our daily routine?.
Answer:
THIS IS NOT MINE.
Step-by-step explanation:
I just search this
25. Directions: For each of the given situational problemsa. Sketch the required triangle to solve the problemb. Use the Law of Cosine to solve the problem
Answer:
Cosine Law:
[tex] {c}^{2} = {a}^{2} + {b}^{2} - (2ab)cos60[/tex]
[tex] {10}^{2} = {a}^{2} + {6}^{2} - (2)(a)(6)cos60[/tex]
[tex]100 = {a}^{2} + 36 - 6a[/tex]
arrange:
[tex] {a}^{2} - 6a - 64 = 0[/tex]
this is a quadratic equation:
[tex]a = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} [/tex]
[tex]a = \frac{ - ( - 6) + \sqrt{ {( - 6)}^{2} - 4(1)( - 64)} }{2(1)} [/tex]
[tex]a = \frac{6 + \sqrt{36 + 256} }{2} [/tex]
[tex]a = \frac{6 + \sqrt{292} }{2} [/tex]
[tex]a = 3 + \sqrt{73} [/tex]
or
a = 11.544 km.26. Given the figure at the right which of the following law can be applied to solve for x?A. Cosine lawB. Heron’s LawC. Pythagorean theoremD. Sine’s Law
A.is the right answer sure yan
hope it help
27. Sketch and solve the following problems using law of cosines or law of sines.
Answer:
Yan..
sana makatulong sayo
28. the law of cosines is use to solve for ______ triangles.
Answer:
the missing longest side and that is hypotenuse
Step-by-step explanation:
C squared = a squared + b squared
29. Applying the law of cosines, solve the following problems. Round to the nearest hundredth.
Answer:
Step-by-step explanation:
Law of Cosine
Used when either the lengths of two sides and the measure of the included angle is given. The lengths of the three sides (SSS) are given. Angle measure should be in degree measure. The measure of the side must be rounded to the nearest hundredth. The sum of the angles should be 180°.
Engagement:
1. Solve ΔABC.
Given: a = ? ∠A = 112°
b = 7 ∠B = ?
c = 12 ∠C = ?
Solve for a. Solve for ∠B.
a² = c² - b² b² = a² + c² - 2ac Cos B
a² = (12)² - (7)² (7)² = (9.75)² + (12)² - 2(9.75)(12) Cos B
a² = 144 - 49 49 = 95 + 144 - 234 Cos B
a² = 95 49 = 239 - 234 Cos B
a = \sqrt{95}
95
49 - 239 = -234 Cos B
a = 9.75 -190 = -234 Cos B
\frac{-190}{-234}
−234
−190
= \frac{-234}{-234}
−234
−234
Cos B
.81 = Cos B
Cos⁻¹ .81 = B
35.90° = B
Solve for ∠C.
∠C = 180° - (∠A + ∠B)
∠C = 180° - (112° + 35.90°)
∠C = 180° - 147.90°)
∠C = 32.10°
2. Solve ΔABC.
Given: a = 42 ∠A = ?
b = 57 ∠B = ?
c = ? ∠C = 28°
Solve for c. Solve for ∠A.
c² = a² + b² a² = b² + c² - 2bc Cos A
c² = (42)² + (57)² (42)² = (57)² + (70.80)² - 2(57)(70.80) Cos A
c² = 1764 + 3249 1764 = 3249 + 5013 - 8071.20 Cos A
c² = 5013 1764 = 8262 - 8071.20 Cos A
c = \sqrt{5013}
5013
1764 - 8262 = -8071.20 Cos A
c = 70.80 -6498 = -80.71.20 Cos A
\frac{-6498}{-8071.20}
−8071.20
−6498
= \frac{-8071.20}{-8071.20}
−8071.20
−8071.20
Cos A
.81 = Cos A
Cos⁻¹.81 = A
35.90° = A
Solve for ∠B.
∠B = 180° - (∠A + ∠C)
∠B = 180° - (35.90° + 28°)
∠B = 180° - 63.90°
∠B = 116.10°
3. Solve ΔABC.
Given: a = 8 ∠A = ?
b = 14 ∠B = ?
c = 17 ∠C = ?
Solve for ∠A. Solve for ∠B.
a² = b² + c² - 2bc Cos A b² = a² + c² - 2ac Cos B
(8)² = (14)² + (17)² - 2(14)(17) Cos A (14)² = (8)² + (17)² - 2(8)(17) Cos B
64 = 196 + 289 - 476 Cos A 196 = 64 + 289 - 272 Cos B
64 = 485 - 476 Cos A 196 = 353 - 272 Cos B
64 - 485 = -476 Cos A 196 - 353 = -272 Cos B
-421 = -476 Cos A -157 = -272 Cos B
\frac{-421}{-476}
−476
−421
= \frac{-476}{-476}
−476
−476
Cos A \frac{-157}{-272}
−272
−157
= \frac{-272}{-272}
−272
−272
Cos B
.88 = Cos A .58 = Cos B
Cos⁻¹.88 = A Cos⁻¹.58 = B
28.36° = A 54.55° = B
Solve for ∠C.
∠C = 180° - (∠A + ∠B)
∠C = 180° - (28.36° + 54,55°)
∠C = 180° - 82.91°
∠C = 97.09°
4. Solve ΔRST.
Given: r = 29 ∠R = ?
s = 16 ∠S = ?
t = 24 ∠T = ?
Solve for ∠R. Solve for ∠S.
r² = s² + t² - 2st Cos R s² = r² + t² - 2rt Cos R
(29)² = (16)² + (24)² - 2(16)(24) Cos R (16)² = (29)² + (24)² - 2(29)(24) Cos S
841 = 256 + 576 - 768 Cos R 256 = 841 + 576 - 1392 Cos S
841 = 832 - 768 Cos R 256 = 1417 - 1392 Cos S
841 - 832 = -768 Cos R 256 - 1417 = -1392 Cos S
9 = -768 Cos R -1161 = -1392 Cos S
\frac{9}{-768}
−768
9
= \frac{-768}{-768}
−768
−768
Cos R \frac{-1161}{-1392}
−1392
−1161
= \frac{-1392}{-1392}
−1392
−1392
Cos S
-.01 = Cos R .83 = Cos S
Cos⁻¹(-.01) = R Cos⁻¹ .83 = S
90.57° = R 33.90° = S
Solve for ∠T.
∠T = 180° - (∠R + ∠S)
∠T = 180° - (90.57° + 33.90°)
∠T = 180° - 124.47°
∠T = 55.53°
exmple lng po to
30. Law of cosine Hope someone can solve this problem.I need it right now please. ty
Answer:
[tex]c 6 \: in \: s \: 9 \: in \: u \: 11 \: in[/tex]
[tex]6 \sqrt{s9u \frac{11 \: u \\ }{?} } [/tex]
