Ask Question

Solve the triangle a = 3, b = 2 and alpha = 40 degrees

posader86 2022-07-30 Answered
Solve the triangle
a = 3, b = 2 and alpha = 40 degrees
Ask Expert 1 See Answers
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

dominicsheq8
Answered 2022-07-31 Author has 15 answers
a / sin ⁡ ( A ) = b / sin ⁡ ( B ) = c / sin ⁡ ( C )
therefore a,b,c are the lengths of the sides
so 3 / sin ⁡ ( 40 d e g ) = 2 / sin ⁡ ( θ ) so arcsin ⁡ ( 2 ∗ sin ⁡ ( 40 ) / 3 ) = B or B=25.37deg
now every triangle is 90deg so 40deg+25.37deg=65.37deg so onlyangle left is 90-65.37=24.63degrees this is opposite of lengthc
so law of sines again using a and c now a / sin ⁡ ( A ) = c / sin ⁡ ( C )so 3 / sin ⁡ ( 40 d e g ) = c / sin ⁡ ( 24.63 d e g ) solve for c=1.945 units in length
the triangle's properties are...
a=3 b=4 c=1.945 A=40deg B=25.37deg C=24.63deg
Not exactly what you’re looking for?
Ask My Question
This is helpful This is helpful 0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2021-09-15

Find where f(x)=4x−tan⁡x, −π2<x<π2 is increasing or decreasing and find it's maximum and minimum values.

See answers (1)
asked 2021-09-10
A right triangle with a side cathetuses equal to 12cm and 5 cm. Find the angle opposite to the cm side.

See answers (1)

Ask your question.
Get your answer.
Easy as that

Ask Question
asked 2021-02-03
Given a triangle with a 112∘ angle. The adjusted sides to this angle are equal. Find their length, if A=80cm2.

See answers (1)
asked 2022-03-29

(ad≥ + 2ycost = coser?t

See answers (0)
asked 2021-11-08
Find  sin⁡2x,cos⁡2x  if  sin⁡x=313  and x terminates in quadrant II

See answers (1)
asked 2021-01-08
PQRS is a parallelogram. m∠Q=11x+13 and m∠S=15x−23. Find the measure of ∠P.

See answers (2)
asked 2021-03-01

Find the exact value of sin⁡60∘.

See answers (1)

New questions

Distance between (-5,3,9) and (-1,-4,1)?
Find all triangles with a fixed base and opposite angle
I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.
I also know the angle opposite the base and this is also fixed.
Now what I want to do is figure out how to compute all possible positions for the third vertex.
My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.
I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis
then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste
now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.
struggling with the maths, do i use the sin rule, i.e. sin a / = sin ⁡ b / B and so on, or do I need to break it down into right angle triangles and then just use something along the lines of a 2 + b 2 = c 2
ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,
base is fixed running from (-1,0) to (1,0) angle opposite the base is 60 deg
I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.
Distance between (-8,1,-3) and (1,4,2)?
Finding the volume of the tetrahedron.
Find the volume of the tetrahedron with the vertices P(1,1,1), Q(1,2,3), R(3,1,2), and S(2,3,1).
In an indirect proof, is it possible to reject the assumption based on contradiction of a premise?
Is there a rule for naming quadrilaterals in English? What I am expected to know about are: square, rhombus, rectangle, parallelogram, trapezium, kite. But how do we name other quadrilaterals?
1). Express 80 inches in standard notation using feet and inches.
2). Find the midpoint between 1.7 and 3.2cm on a metric rule.
Lebesgue measure using special polygons
I want to prove that the following set satisfies m ( G ) = 1.
G = { ( x , y ) ∈ R 2 : x > 0 , 0 < y < e − x }
I have to prove this without using the Lebesgue integral.
This set is open, so its measure is defined as it follows:
m ( G ) = sup { m ( P ) : P ⊂ G }, where P is a special polygon.
We can construct a sequence of special polygons P n ⊂ G:
P n = ⋃ j = 1 n [ 0 , l o g ( n + 1 j ) ] × [ j − 1 n + 1 , j n + 1 ] .
The measure of one of these special polygons is m ( P n ) = ∑ j = 1 n l o g ( n + 1 j ) 1 n + 1 .
As the sequence of measures { m ( P n ) } is increasing, the supremum of this sequence is its limit:
sup ( { m ( P n ) } ) = lim n → ∞ ∑ j = 1 n l o g ( n + 1 j ) 1 n + 1 = lim n → ∞ log ⁡ ( ( n + 1 ) n n ! ) 1 n + 1 .
I know that this limit is 1, so the logarithm argument must tend to e, but I don't know how to prove this analytically. I've tried working on the final expression to get the definition of e, but I haven't been able to do so. Could yo help me?
Distance between (-10,-2,2) and (4,-4,-2)?
Show that the proposition P:

There exists a pair of straight lines that are at constant distance from each other.

is equivalent to the Parallel Postulate Q :

If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough.

I tried to prove Q → P then ¬ Q → ¬ P. But for the second part, I can do nothing because as soon as the postulate is supposed to be untrue, the equivalent relation between angles no more exist, therefore it's hard to get congruent triangles as I used to do.

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
PlainMath Secondary Post Secondary
Secured by PayPal facebook youtube
About Us Term of Service Privacy Contact Us Honor Math Solver 2022 Plainmath. All rights reserved
Company info