Exercises: Using trig ratios to find missing sides and angles - Digital SAT Math

A

Recall / Warm-Up

1.

In the right triangle below, angle $A = 40\degree$ . If you know the hypotenuse and want to find the side opposite angle $A$ , which trig ratio should you use?

A.

Cosine

B.

Tangent

C.

Sine

D.

It doesn't matter — any ratio will work

2.

Which equation correctly defines $\\tan(A)$ in a right triangle?

A.

$\\tan(A) = \\dfrac{\\text{hypotenuse}}{\\text{adjacent}}$

B.

$\\tan(A) = \\dfrac{\\text{opposite}}{\\text{hypotenuse}}$

C.

$\\tan(A) = \\dfrac{\\text{opposite}}{\\text{adjacent}}$

D.

$\\tan(A) = \\dfrac{\\text{adjacent}}{\\text{opposite}}$

3.

A right triangle has legs 3 and 4 and hypotenuse 5. Angle $A$ is the angle opposite the side of length 3.
What is $\\sin(A)$ ? Enter your answer as a fraction or decimal.

B

Fluency Practice

1.

In a right triangle, angle $B = 52\degree$ and the side adjacent to $B$ measures 15. To find the side opposite $B$ , which equation is correct?

A.

$\\sin(52\degree) = \\dfrac{\\text{opposite}}{15}$

B.

$\\cos(52\degree) = \\dfrac{\\text{opposite}}{15}$

C.

$\\tan(52\degree) = \\dfrac{\\text{opposite}}{15}$

D.

$\\tan(52\degree) = \\dfrac{15}{\\text{opposite}}$

2.

A right triangle has angle $A = 35\degree$ and hypotenuse $= 20$ . Find the length of the side opposite angle $A$ . Round to the nearest hundredth.

3.

A right triangle has angle $B = 60\degree$ and the side adjacent to $B$ measures 10. Find the hypotenuse.

4.

A right triangle has angle $C = 45\degree$ and the side opposite to $C$ measures 9. Find the adjacent side.

5.

In a right triangle, the side opposite angle $A$ measures 8 and the hypotenuse is 17. Use the inverse sine function to find angle $A$ to the nearest degree.

C

Varied Practice

1.

A right triangle has angle $P = 30\degree$ and hypotenuse $= 14$ . Which expression gives the length of the side adjacent to $P$ ?

A.

$14 \\cdot \\sin(30\degree)$

B.

$14 \\cdot \\tan(30\degree)$

C.

$14 \\cdot \\cos(30\degree)$

D.

$14 / \\sin(30\degree)$

2.

A right triangle has legs of length 9 and 12. Find the acute angle opposite the side of length 9, to the nearest degree.

3.

In a right triangle, angle $Q = 25\degree$ and the side opposite $Q$ measures 7. Find the adjacent side. Round to the nearest tenth.

4.

To find angle $A$ in a right triangle when you know two sides:

Given opposite $= 5$ and hypotenuse $= 13$ , first write the ratio:

$\\sin(A) =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ / ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲

Then apply the inverse function:

$A =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $^{-1}$ ( ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ / ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ )

numerator:

denominator:

function name:

numerator (inverse):

denominator (inverse):

5.

A ramp makes a $15\degree$ angle with the horizontal. The horizontal distance covered by the ramp is 30 feet.

What is the vertical rise of the ramp? Round to the nearest tenth of a foot.

D

Word Problems / Application

1.

A 20-foot flagpole stands vertically on level ground. The angle of elevation from the tip of the shadow to the top of the pole is $55\degree$ .

How long is the shadow, in feet? Round to the nearest foot.

2.

Two sides of a right triangle are known: the side adjacent to angle $\\theta$ measures 7 cm, and the hypotenuse measures 25 cm.

Find angle $\\theta$ to the nearest degree.

3.

A person stands 50 meters from the base of a tower on level ground and looks up at the top of the tower. The angle of elevation to the top is $40\degree$ .

1.

How tall is the tower, in meters? Round to the nearest tenth.

2.

If the angle of elevation were increased from $40\degree$ to $60\degree$ (while staying 50 meters from the base), what would happen to the tower height calculated?

A.

It would decrease.

B.

It would stay the same.

C.

It would increase.

D.

Cannot be determined without more information.

4.

From the top of a 200-foot cliff, a rescue team observes a boat at sea. The angle of depression from the top of the cliff to the boat is $25\degree$ .

How far is the boat from the base of the cliff, in feet? Round to the nearest foot.

E

Error Analysis

1.

Alex solved this problem:

"In a right triangle with angle $A = 40\degree$ and adjacent side $= 12$ , find the side opposite $A$ ."

Alex's work:
$\\sin(40\degree) = \\dfrac{12}{x}$ , so $x = \\dfrac{12}{\\sin(40\degree)} \\approx 18.67$

What mistake did Alex make?

A.

Alex should have used cosine instead of sine.

B.

Alex chose the wrong trig ratio — tangent connects the opposite and adjacent sides. The correct equation is $\\tan(40\degree) = x/12$ , giving $x = 12\\tan(40\degree) \\approx 10.07$ .

C.

Alex set up the fraction upside down.

D.

The adjacent side should be in the numerator of the sine ratio.

2.

Jordan solved this problem:

"Find angle $A$ if the opposite side $= 5$ and the hypotenuse $= 10$ ."

Jordan's work:
$A = \\sin\\left(\\dfrac{5}{10}\\right) = \\sin(0.5) \\approx 28.6\degree$

What error did Jordan make?

A.

Jordan used the wrong trig ratio — should have used cosine.

B.

Jordan confused trig and inverse trig — should use $\\sin^{-1}(5/10) = \\sin^{-1}(0.5) = 30\degree$ , not $\\sin(5/10)$ .

C.

Jordan made an arithmetic error in the division.

D.

The answer of $28.6\degree$ is actually correct.

F

Challenge / Extension

1.

A ladder 15 feet long leans against a vertical wall. The base of the ladder makes a $65\degree$ angle with the ground.

Find the height that the top of the ladder reaches up the wall. Round to the nearest tenth of a foot.

2.

A surveyor at point $A$ measures an angle of elevation of $30\degree$ to the top of a hill.
The surveyor then walks 100 meters directly toward the hill to point $B$ , where the
angle of elevation to the top of the hill is $60\degree$ .

1.

Let $h$ be the height of the hill and $d$ be the horizontal distance from point $B$ to the base of the hill. From point B, which equation correctly relates $h$ and $d$ ?

A.

$h = d \\cdot \\tan(60\degree)$

B.

$h = d \\cdot \\cos(60\degree)$

C.

$h = d / \\tan(60\degree)$

D.

$h = (d + 100) \\cdot \\tan(30\degree)$

2.