Practice Problems-Amoagh Gopinath
1. Solve for x,y, and z.
Solving for x:
1. First set up a proportion to find the geometric mean(x) of 5.4 and 5.4+9.6
5.4/x=x/5.4+9.6
2.Cross multiply
x^2=81
3.Solve for x
x=9
Solving for y:
1.Set up Pythagorean theorem because it's a right triangle and you're give two sides.
9^2+y^2=15^2
2.solve for y
81+y^2=225
y^2=144
y=12
Solving for z:
1.Use heartbeat theorem to find z. Set up proportion
5.4/z=z/9.6
2.cross multiply
z^2=51.84
3.Solve for z
z=7.2
2.Find the area and perimeter of the regular hexagon. Round to the nearest hundredth if necessary.
1. First set up a proportion to find the geometric mean(x) of 5.4 and 5.4+9.6
5.4/x=x/5.4+9.6
2.Cross multiply
x^2=81
3.Solve for x
x=9
Solving for y:
1.Set up Pythagorean theorem because it's a right triangle and you're give two sides.
9^2+y^2=15^2
2.solve for y
81+y^2=225
y^2=144
y=12
Solving for z:
1.Use heartbeat theorem to find z. Set up proportion
5.4/z=z/9.6
2.cross multiply
z^2=51.84
3.Solve for z
z=7.2
2.Find the area and perimeter of the regular hexagon. Round to the nearest hundredth if necessary.
Perimeter:
1.Find the measure of each angle.
180(6-2)/6=120
2.Find the measure of half of an angle.
120/2=60
3.Set up a function to find half of the side length and solve.
7/x=tan60
7/tan60=x
x=4.0415
4.Find side length then perimeter.
4.0415*2=8.083
8.083*6=48.5
Perimeter=48.5 in.
Area:
A=1/2aP
A=1/2*7*48.5
A=169.75 in squared
1.Find the measure of each angle.
180(6-2)/6=120
2.Find the measure of half of an angle.
120/2=60
3.Set up a function to find half of the side length and solve.
7/x=tan60
7/tan60=x
x=4.0415
4.Find side length then perimeter.
4.0415*2=8.083
8.083*6=48.5
Perimeter=48.5 in.
Area:
A=1/2aP
A=1/2*7*48.5
A=169.75 in squared
Practice Problem - Sai
Directions: do the following proof
Given: triangle ABC with DE || BC
Prove: BD/DA = CE/EA
Prove: BD/DA = CE/EA
Explanation: Since we are given that segment DE is parallel to BC(forms 2 triangles), we then have a starting point. We need to prove similarity for both triangles and we can do that because of the parallel lines and corresponding angles theorem. Then we have similarity through Angle-Angle similarity. Now to prove the statement we need to manipulate what we know from the similarity of the 2 triangles. We can first set up the proportion to compare the sides because we know the sides are proportional(triangles proportionality theorem) and then we can manipulate this statement by using the segment addition postulate to change this statement. Then when we use the subtraction property, we can end up with the statement that we needed to prove.
Practice Problem - Sai
Directions: solve for the missing lengths using trig ratios, then solve using the 30-60-90 special triangle rule and see if answers match(round to the nearest whole number), then find the hypotenuse using a trig function and check using pythagorean theorem
Answer/Explanation: To solve this problem, we need to solve one variable and then use that variable to solve the other. Then we can check our answers using the special triangle rule. So first, let's solve for r because we are given enough pieces of information to solve for that length. Using the 30 degree angle as our beta angle, we can use the trig function tangent because we have our opposite as r, our missing length that we are solving or and 60 which is the adjacent angle. Equation: tan 30 = r / 60
tan30(60) = r r = 34.6410161514 or 35
We found the value of r and we can than use this value to solve for s in the smaller triangle. Using the 55 degree angle as our beta angle,
we can use the same trig function tangent because we are given r as our opposite and s as our adjacent and we just need to substitute
the solved value for r. Equation: tan 55 = 35 / s s = 35 / (tan55)
s = 24.5072638373 or 25
We found the value of r which is 35 and s which is 25. Now let's check using the 30-60-90 rule.
60 / sqrt3 = 20sqrt3 Side r = 20sqrt3 = 35 Side r is correct according to 30-60-90 rule. We can't check S but because r is solved for the same way s was, we can assume it's correct.
Not let's solve for the hypotenuse using a trig function(any would work because we have all the information we need plus more) and then check using the pythagorean theorem.
We can use the 30 degree base angle to solve for the hypotenuse. We can use cosine because we can use 60 as the adjacent and x as our hypotenuse.
Equation: cosine 30 = 60 / x (hyp.) x = 60 / cos 30 x = 69.28 or 69
Using trig functions, we got about 69 as our hypotenuse and we can now use pythagorean theorem to check.
35 ^2 + 60 ^2 = 1225 + 3600 = sqrt 4825 = 69.46 or 69
We checked all our answers using older methods like 30-60-90 and pythagorean theorem and we still got the same answer as using trig functions.
tan30(60) = r r = 34.6410161514 or 35
We found the value of r and we can than use this value to solve for s in the smaller triangle. Using the 55 degree angle as our beta angle,
we can use the same trig function tangent because we are given r as our opposite and s as our adjacent and we just need to substitute
the solved value for r. Equation: tan 55 = 35 / s s = 35 / (tan55)
s = 24.5072638373 or 25
We found the value of r which is 35 and s which is 25. Now let's check using the 30-60-90 rule.
60 / sqrt3 = 20sqrt3 Side r = 20sqrt3 = 35 Side r is correct according to 30-60-90 rule. We can't check S but because r is solved for the same way s was, we can assume it's correct.
Not let's solve for the hypotenuse using a trig function(any would work because we have all the information we need plus more) and then check using the pythagorean theorem.
We can use the 30 degree base angle to solve for the hypotenuse. We can use cosine because we can use 60 as the adjacent and x as our hypotenuse.
Equation: cosine 30 = 60 / x (hyp.) x = 60 / cos 30 x = 69.28 or 69
Using trig functions, we got about 69 as our hypotenuse and we can now use pythagorean theorem to check.
35 ^2 + 60 ^2 = 1225 + 3600 = sqrt 4825 = 69.46 or 69
We checked all our answers using older methods like 30-60-90 and pythagorean theorem and we still got the same answer as using trig functions.