Home / Digital SAT Math Practice Questions -Medium : Right triangles and trigonometry

# Digital SAT Math Practice Questions -Medium : Right triangles and trigonometry

## SAT MAth Practice questions – all topics

• Geometry and Trigonometry Weightage: 15% Questions: 5-7
• Area and volume
• Lines, angles, and triangles
• Right triangles and trigonometry
• Circles

## SAT MAth and English  – full syllabus practice tests

[Calc]  Question  medium

For the triangle shown, which equation is NOT true?
A. $$\sin x=\frac{42}{59}$$
B. $$\sin \left(90^{\circ}-x\right)=\frac{42}{59}$$
C. $$\cos \left(90^{\circ}-x\right)=\frac{42}{59}$$
D. $$\sin \left(90^{\circ}-x\right)-\cos x=0$$

Ans:B

Given $$\sin x =\frac{\text{Perpendicular}}{\text{Hypot.}}= \frac{42}{59}$$, we know that in a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

$$\sin \left(90^{\circ}-x\right)=cos x$$

From triangle , $$\cos x =\frac{\text{base}}{\text{Hypot.}}$$

$$\cos x =\frac{\text{41.4}}{\text{59}}$$ So option- B is wrong

[No calc]  Question   medium

An isosceles right triangle has a hypotenuse of length 4 inches. What is the perimeter, in inches, of this triangle?

A. $$2 \sqrt{2}$$
B. $$4 \sqrt{2}$$
C. $$4+4 \sqrt{2}$$
D. $$4+8 \sqrt{2}$$

Ans:C

An isosceles right triangle has two equal sides, each of which is the same length. In this case, the hypotenuse is given as 4 inches.

By the Pythagorean theorem, if $$a$$ and $$b$$ are the lengths of the two legs of the right triangle, and $$c$$ is the length of the hypotenuse, then $$a^2 + b^2 = c^2$$.

Since it’s an isosceles right triangle, the legs are equal. Let’s call the length of each leg $$x$$.

So, $$x^2 + x^2 = 4^2$$.

Solving for $$x$$:
$2x^2 = 16$
$x^2 = 8$
$x = \sqrt{8}$
$x = 2\sqrt{2}$

The perimeter of the triangle is the sum of all three sides:

$P = 2x + 4$

Substituting the value of $$x$$:

$P = 2(2\sqrt{2}) + 4$
$P = 4\sqrt{2} + 4$

So, the correct answer is option C: $$4 + 4\sqrt{2}$$.

[Calc]  Question  medium

Rectangle X has a length of 12 centimeters (cm) and a width of 2.5 cm. Right triangle Y has a base of 10 cm. The area of rectangle X is three times the area of right triangle Y. What is the height, in cm, of right triangle Y?
A. 1
B. 2
C. 3
D. 6

Ans: B

The area of a rectangle is given by the formula $$\text{Area} = \text{length} \times \text{width}$$. Similarly, the area of a right triangle is given by $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$.

Given that the area of rectangle $$X$$ is three times the area of right triangle $$Y$$, we can set up the equation:

$12 \times 2.5 = 3 \times \frac{1}{2} \times 10 \times \text{height of triangle } Y$

$30 = 15 \times \text{height of triangle } Y$

$\text{height of triangle } Y = \frac{30}{15} = 2$

So, the height of right triangle $$Y$$ is $$2$$ centimeters. Therefore, the correct answer is option B, $$2$$.

[Calc]  Question  medium

Which of the following additional pieces of information provides enough information to prove whether triangle DEF is a right triangle?
I. The measure of angle D
II. The length of segment DF
A. I only
B. II only
C. Either I or II
D. Neither I nor II

Ans: C

To determine if a triangle is a right triangle, we need to either know the measure of one of its angles being 90 degrees, or use the Pythagorean theorem to check if the sum of the squares of the lengths of two sides equals the square of the length of the third side.

• The diagram shows a triangle DEF with side lengths DE = 3 units and EF = 5 units.

I. The measure of angle D:

If the measure of angle D is 90 degrees, then triangle DEF is definitely a right triangle.

If the measure of angle D is not 90 degrees, then we can determine that triangle DEF is not a right triangle.

II. The length of segment DF:

If we know the length of segment DF, we can use the Pythagorean theorem to check if $\rm{DF^2 = DE^2 + EF^2}$. If this equation holds true, then the triangle is a right triangle.

Therefore, the correct answer should be C. Either I or II

[Calc]  Questions  Medium

Triangle ABC is similar to triangle DEF, where angle A corresponds to angle D. What is the value of cos F ?

Ans: 4/5, .8

Since the triangles are similar, their corresponding angles are equal. Therefore, we have: Angle A $$=$$ Angle $$D$$

From the given figure, we can observe that angle $$\mathrm{F}$$ is the complement of angle $$\mathrm{A}$$ (or angle D).

Angle $$\mathrm{F}=90^{\circ}$$ – Angle A
To find the value of $$\cos \mathrm{F}$$, we need to know the value of angle $$\mathrm{F}$$ in degrees.

Using the trigonometric ratio for cosine:
$\cos \mathrm{F}=\cos \left(90^{\circ}-\mathrm{A}\right)=\sin \mathrm{A}$

\begin{aligned} \operatorname{Sin A} & =\frac{\text { Perpend. }}{\text { Hypot. }} \\ & \operatorname{Sin A}=\frac{24}{30} \\ & \operatorname{Sin A}=\frac{4}{5}\end{aligned}

[No- Calc]  Question Medium

In the right triangle $$\mathrm{PQR}$$, the length of side $$\overline{\mathrm{PQ}}$$ is 70 , the measure of angle $$\mathrm{P}$$ is $$90^{\circ}$$, and the measure of angle $$\mathrm{R}$$ is $$38^{\circ}$$. Which of the following represents the length of side $$\overline{\mathrm{QR}}$$ ?
A) $$\frac{70}{\sin 52^{\circ}}$$
B) $$\frac{70}{\sin 38^{\circ}}$$
C) $$70 \sin 52^{\circ}$$
D) $$70 \sin 38^{\circ}$$

Ans:B

In a right triangle, the ratio of the side opposite an angle to the hypotenuse is given by the sine function.

Solution: According to sine law
\begin{aligned} & \Rightarrow \frac{P Q}{\sin R}=\frac{R Q}{\sin P}=\frac{P R}{\sin \theta} \\ & \Rightarrow \frac{P Q}{\sin R}=\frac{R Q}{\sin P} \Rightarrow \frac{70}{\sin 38}=\frac{R Q}{\sin 90} . \\ & \Rightarrow R Q=\frac{70 \times \sin 90}{\sin 38^{\circ}}=\frac{70 \times 1}{\sin 38}=\frac{70}{\sin 38^{\circ}} \end{aligned}

Hence $$R Q=70 / \sin 38^{\circ}$$

Question

Triangle $$DEF$$ (not shown) is similar to triangle $$ABC$$ above, where side $$DE$$corresponds to side $$AB$$, side $$DF$$ corresponds to side $$AC$$, and $$DE$$ = 2$$AB$$. What is the measure of angle $$DFE$$  ?

1. 37°
2. 53°
3. 74°
4. 106°

B

Question

What is the length of side AC in the triangle below?

1. $5\sqrt{3}$
2. 10
3. 15
4. $10\sqrt{3}$

B

Question

In the $x y$-plane above points $P, Q, R$, and $T$ lie on the circle with center $O$. The degree measures of angles $P O Q$ and $R O T$ are each $30^{\circ}$. What is the radian measure of the angle $Q O R$ ?
A. $\frac{5}{6} \pi$
B. $\frac{3}{4} \pi$
C. $\frac{2}{3} \pi$
D. $\frac{1}{3} \pi$

Ans: C

Questions

In a right triangle, the tangent of one of the two acute angles is $\frac{\sqrt{3}}{3}$. What is the tangent of the other acute angle?
A. $-\frac{\sqrt{3}}{3}$
B. $-\frac{3}{\sqrt{3}}$
C. $\frac{\sqrt{3}}{3}$
D. $\frac{3}{\sqrt{3}}$

Ans: D

Questions

In the right triangle above, $x=60$. What is the length of side $\overline{A B}$ ?
A. 7
B. 8
C. 9
D. It cannot be determined from the information given.

Ans: B

Questions

In the figure above, △$$ACD$$ is a right triangle and $$\overline{BE}$$ is parallel to $$\overline{CD}$$. What is the perimeter of △$$ACD$$ to the nearest tenth of a unit?

1. 29.7
2. 36.0
3. 41.5
4. 50.9

Ans: C

Question

In the figure above, line $$l$$. is parallel to line $$m$$. If $$x = 40$$, what is the measure of ∠DEF ?

1. 140°
2. 100°
3. 80°
4. 50°

Ans: B

Questions

.

In the figure above, $$sin(90° – x° ) = \frac{12}{13}$$. What is the value of $$sin x°$$ ?

1. $$\frac{12}{13}$$
2. $$\frac{5}{13}$$
3. $$\frac{5}{12}$$
4. $$\frac{13}{12}$$

Ans: B

Questions

Triangles $$ABC$$ and $$DEF$$ above are similar. How much longer than segment $$EF$$ is segment $$DE$$?

1. 1
2. 2
3. 4
4. 8