Question 1206928: Sketch a figure and solve each right triangle ABC with right angle at C. given that:
1. A = 15° and c = 37
2. B = 64° and c = 19.2
3. A = 15° and c = 25
4. A = 45° and c = 16
5. B = 56° and c = 16
Found 6 solutions by MathLover1, ikleyn, josgarithmetic, math_tutor2020, MathTherapy, mccravyedwin:
Answer by MathLover1(20850)   (Show Source):
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
Sketch a figure and solve each right triangle ABC with right angle at C. given that:
1. A = 15° and c = 37
2. B = 64° and c = 19.2
3. A = 15° and c = 25
4. A = 45° and c = 16
5. B = 56° and c = 16
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In this problem, in each case you are given a right angled triangle,
one acute angle and the hypotenuse length.
In each case, your first step is to find the other acute angle as a complementary angle
to the given acute angle. After that, you know BOTH acute angles in your triangle.
Your next step is to find the legs of the right-angled triangle.
You do it by multiplying the hypotenuse by sine of the opposite acute angle.
This is all the necessary methodology.
How @MathLover1 teaches you in her post, is ANTI-pedagogy in its finished form.
Answer by josgarithmetic(39630)  (Show Source):
You can put this solution on YOUR website! #3
Enough to be able to draw the triangle and label what is given.
Sum of Interior Angles of Triangle means B is 75 degrees.
Simple Trigonometry:
 ;
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
The Law of Sines is a bit overkill, but it is a method you could follow.
A more easier approach would be to use the SOH-CAH-TOA trig ratios- SOH = sine opposite hypotenuse
- CAH = cosine adjacent hypotenuse
- TOA = tangent opposite adjacent
I'll focus on problem 1 only and leave the others for the student.
C = 90 degrees
A = 15 degrees
B = 90-A = 90-15 = 75 degrees
Angles A and B are complementary, i.e. they add to 90 degrees, since we're dealing with a right triangle.
Now that we know the angles, we can determine the side lengths.
sin(angle) = opposite/hypotenuse
sin(A) = a/c
sin(15) = a/37
a = 37*sin(15)
a = 9.576305 approximately
Please make sure that your calculator is set to degrees mode.
I often check by typing in something like "sin(30)" and the output for this example should be 0.5
And,
sin(angle) = opposite/hypotenuse
sin(B) = b/c
sin(75) = b/37
b = 37*sin(75)
b = 35.739256 approximately
--------------------------------------------------------------------------
Here is the fully solved triangle for problem 1
Angles: A = 15, B = 75, C = 90
Sides: a = 9.576305 approximately, b = 35.739256 approximately, c = 37
Round the approximate values however needed.
I'll leave the sketch for the student to do, as well as problems 2 through 5.
More practice found here
https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1202430.html
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
Sketch a figure and solve each right triangle ABC with right angle at C. given that:
1. A = 15° and c = 37
2. B = 64° and c = 19.2
3. A = 15° and c = 25
4. A = 45° and c = 16
5. B = 56° and c = 16
Law of Sines? Totally UNNECESSARY here.
1. A = 15° and c = 37
GIVEN: A triangle with angle A being 15o, hypotenuse AB (c) being 37 units, OPPOSITE
side, BC (a), and ADJACENT side AC (b).
Side b (AC) = 37 * cos (15o) = 35.74 units ------ Cross-multiplying
Needless to say, Acute angle B = 75o (90o - 15o)
Side a (BC) = 37 * cos (90 - 15)o = 37 * cos (75o) = 9.58 units.
Apply same concept to the rest, sketching EACH triangle as you go along!!
Answer by mccravyedwin(409)  (Show Source):
You can put this solution on YOUR website!
Here are the only two formulas you need to solve any right triangle ABC with
right angle at C when given A or B, and c.
Armed with those and a scientific calculator, you don't need a tutor.
I could give you formulas for when you're given other parts, but you
don't have any of those above. When you get to those, post again and I'll
give you the formulas for them, too.
Edwin
|
|
|
