Download 3B Review Questions continued.

Geometry
Unit 3B Review
3B Test on Monday Feb 6
Special Right Triangles
2n
n
n√2
n√3
n
n
BC is side OPPOSITE the reference angle
CA is the side ADJACENT to the reference angle
B
sinθ°= Opp
hypot
cosθ°= Adj
hypot
tanθ°= opp
adj
opp
C
hypot
adj
θ° A
3B Review - Team Test
1. Find geometric mean between “a” and “b” (any
2 numbers). Give an exact answer.
b
x
a
=
x2 = ab
x
x = √ab
Simplify as much as you can,
but OK to have the radical - is
“An exact answer.”
2. Determine the values for x, y and z. Give exact
answers in simplified radical form.
Two similar triangles - so
corresponding sides are
proportional
x
y
7
z
16
y 16
=
7 y
y2 = 7(16)
y = 4√7
2. Determine the values for x, y and z. Give exact
answers in simplified radical form.
x
y
7
z
16
Then use pythagorean
theorem
x2 = 72 + (4√7)2
= 49 +112 = 161
x = √161
2. Determine the values for x, y and z. Give exact
answers in simplified radical form.
x
y
7
z
16
Then use pythagorean
theorem
z2 = 162 + (4√7)2
= 256 +112 = 368
z = √16(23)
z = 4√23
2. 115 feet tall lighthouse, plus 45ft land
18°
160 ft total
height above
sea level
d
Sin 18° = 160 ft
d
d = 160 ft
0.3090
d -- 518ft
3. Find x.
A. cos 16° = sin x
When cosine of an angle is equal to the sine of
another angle it is because the two angles are
complementary angles - add up to 90°
So
16° + x = 90°
x = 74°
3. Find x.
B. cos (2x + 15)° = sin (x + 9)
When cosine of an angle is equal to the sine of
another angle it is because the two angles are
complementary angles - add up to 90°
So
2x + 15 + x + 9 = 90°
3x + 24 = 90° 3x = 66°
x = 22°
4. Find x and y. Exact answers only must be a special right triangle
question.
30-60-90 triangle
14 cm = y√3
30°
y
x = 2y
Here
y = 14cm
√3
But we shouldn’t leave the
answer with a radical sign
in the denominator
4. Find x and y. Exact answers only must be a special right triangle
question.
30-60-90 triangle
14 cm = y√3
30°
y
x = 2y
Here
y = 14cm √3
√3 √3
y = 14√3
3
4. Find x and y. Exact answers only must be a special right triangle
question.
then
14 cm = y√3
30°
y
x = 2y
x = 2(14√3 )
3
x = 28√3
3
Check your answer is
simplified.
5. Find x and y. Exact answers only must be a special right triangle
question.
x = 8√3 cm
x = base√3
30°
8cm
y = 2 base
length
y = 2(8cm) = 16cm
6. Find x.
45-45-90 triangle
18cm = x√2
x
x = 18 √2
√2 √2
45°
18 cm
x
x = 9√2 cm
rationalize the denominator
7. Find x.
45-45-90 triangle
25√2cm = x√2 cm
x = 25√2 cm
√2
45°
25√2 cm
x
x = 25 cm
8. Solve for the side x. Round all final
answers to 2 decimal places. Use
calculator.
x
28cm
50°
The side x is adjacent to the
reference angle, 50°
Need to use cosine
Cos50° = adjacent = x
Hypotenuse 28cm
0.6428 (28cm) = x
Look up cosine of 50 degrees in the cosine table
you have - then use calculator to find answer round to 2 decimal places
8. Solve for the side x. Round all final
answers to 2 decimal places. Use
calculator.
x
28cm
50°
The side x is adjacent to the
reference angle, 50°
Need to use cosine
Cos50° = adjacent = x
Hypotenuse 28cm
0.6428 (28cm) = x = 17.9984
x approx. 18.00 cm
9. Find the missing angle. Round to
nearest tenths. Use calculator.
θ°
44cm
The 32cm side opposite to the
reference angle, θ°
Need to use sine
sinθ° = opposite = 32 cm
32cm
Hypotenuse 44cm
Sinθ° = (0.72)
θ° = sin-1(0.72) approx. 46.7°
9. Find the missing angle. Round to
nearest tenths. Use calculator.
θ°
Check your answer
makes sense using
the sine table you
have 44cm
- you may punch
in the numbers on the
calculator and make a
mistake doing that
The 32cm side opposite to the
reference angle, θ°
Need to use sine
sinθ° = opposite = 32 cm
32cm
Hypotenuse 44cm
Sinθ° = (0.72)
θ° = sin-1(0.72) approx. 46.7°
10. Which trigonometric ratios are
equivalent to ⅗. Select 2 that apply.
6
P
Q
θ°
8
10
R
sinθ° = opposite
hypotenuse
cosθ° = adjacent
Hypotenuse
tanθ° =opposite =
Adjacent
NOT FINISHED...
=8
10
=6
10
8
6
10. Which trigonometric ratios are
equivalent to ⅗. Select 2 that apply.
6
Q
P
8
10
a°
R
sina° = opposite
hypotenuse
cosa° = adjacent
Hypotenuse
tanθ° =opposite =
Adjacent
=6
10
=8
10
6
8
11. Find x.
Pythagorean theorem
c2 = a 2 + b 2
5
4
52 = 4 2 + x 2
25 - 16 = x2
x
x = √9 = 3
12. 25ft ladder, 65° angle with ground. How far
from the base of the building is the foot of the
ladder. Round to tenths place.
cos 65° = x
25ft
x = 25ft(cos 65°)
25ft
x -- 25ft(0.4226)
x -- 10.6 ft
65°
x
13. A 4.3ft guy wire is attached to a tree,
4ft from the ground. What is the angle
that is formed between the wire and the
ground, to the nearest degree?
sin θ° = 4ft
4.3ft
θ°= sin-1( 0.9302)
4ft is the side
opposite the
reference angle, and
4.3ft is the
hypotenuse - so need
to use sine
relationship
4.3ft
θ° - 68°
θ°
4ft
14. Flying a kite. String forms an angle of
elevation of 27° and distance to directly
under the kite is 35 ft. How long is the kite
string to the nearest foot.
We know the side
cos 27° = 35 ft
xft
x = 35ft
0.8910
x - 39 ft
x
27°
35ft
length that is adjacent
to the reference angle,
and we know the angle
itself, so we can use
cosine of the angle
and the side length
given to calculate the
length of the string
(hypotenuse of the
right triangle.)