SOLUTION: sketch q figure and solve each right triangle ABC with right angle at C.
1. A = 76° and a = 13
2. A = 22° and b = 22
3. B = 30° and b = 11
4. B = 18° and a = 18
5. A = 77Â
Algebra ->
Trigonometry-basics
-> SOLUTION: sketch q figure and solve each right triangle ABC with right angle at C.
1. A = 76° and a = 13
2. A = 22° and b = 22
3. B = 30° and b = 11
4. B = 18° and a = 18
5. A = 77Â
Log On
Question 1202430: sketch q figure and solve each right triangle ABC with right angle at C.
1. A = 76° and a = 13
2. A = 22° and b = 22
3. B = 30° and b = 11
4. B = 18° and a = 18
5. A = 77° and b = 42
Only commenting or helping #3:
Draw it! You have a 30-60-90 right triangle. You know this comes from half of a equilateral triangle. Therefore you will figure that c is two times b; so c=22. IF necessary (may not be depending what you know), Pythagorean Theorem will give you length a. You already know angle at A is 60 degrees.
The phrasing "solve the triangle" means "find the missing sides and missing angles".
Side a is opposite angle A.
Side b is opposite angle B.
Side c is opposite angle C.
The right angle, aka 90 degree angle, is at C
C = 90
We denote this with a square angle marker as shown in the diagrams below.
Angles A and B are complementary. They add to 90
A+B = 90
76+B = 90
B = 90-76
B = 14
Or we could say:
A+B+C = 180
76+B+90 = 180
166+B = 180
B = 180-166
B = 14
This is what we have so far
We'll use the soh cah toa trig ratios to fill in the missing sides.
soh = sine opposite hypotenuse
cah = cosine adjacent hypotenuse
toa = tangent opposite adjacent
Here is one way to find side c.
sin(angle) = opposite/hypotenuse
sin(A) = a/c
sin(76) = 13/c
c*sin(76) = 13
c = 13/sin(76)
c = 13.39797718
c = 13.398 which is approximate
Make sure your calculator is in degree mode.
I'll let the student explore the 2nd pathway to determine c (hint: cos(B) = a/c).
Let's find side b.
sin(angle) = opposite/hypotenuse
sin(B) = b/c
sin(14) = b/13.39797718
b = 13.39797718*sin(14)
b = 3.24126404
b = 3.241
There are three more pathways using a trig ratio, which I'll let the student explore further (hints: cos(A) = b/c, tan(A) = a/b, and tan(B) = b/a)
Or we could use the pythagorean theorem
a^2+b^2 = c^2
13^2 + b^2 = (13.39797718)^2
169 + b^2 = 179.5057925158
b^2 = 179.5057925158 - 169
b^2 = 10.5057925158
b = sqrt(10.5057925158)
b = 3.24126403055969
b = 3.241
Here's the completed right triangle with all angles and all sides labeled. The decimal values are approximate. Round however your teacher instructs.