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
In right triangle \(ABC\), angle \(C\) is a right angle and \(\sin A = 0.70\). What is the value of \(\cos B\) ?
A) 0.60
B) 0.65
C) 0.70
D) 0.75
▶️ Answer/Explanation
Answer: 0.70 or \(\frac{7}{10}\)
In right triangle \(ABC\) with angle \(C\) as the right angle, \(\sin A = 0.70\). Since \(\angle A\) and \(\angle B\) are complementary, \(\cos B = \sin A = 0.70\).
In triangle DEF, point G (not shown) lies on \(\bar{DE}\). If the measure of \(\angle DFG\) is \(x^{\circ}\) and the measure of \(\angle GFE\) is \(y^{\circ}\), what is the value of \(\cos x^{\circ} – \sin y^{\circ}\)?
A) -0.5
B) 0
C) 0.5
D) 1
▶️ Answer/Explanation
Answer: 0
\(x^{\circ} + y^{\circ} = 90^{\circ}\), so \(x^{\circ} = 90 – y^{\circ}\). Then, \(\cos x^{\circ} = \cos (90 – y^{\circ}) = \sin y^{\circ}\), thus \(\cos x^{\circ} – \sin y^{\circ} = 0\).
Question Hard
What is the value of \(\sin(\frac{3\pi}{4})\)
A) \(\frac{-\sqrt{2}}{2}\)
B) \(\frac{-\sqrt{3}}{2}\)
C) \(\frac{\sqrt{2}}{2}\)
D) \(\frac{\sqrt{3}}{2}\)
▶️Answer/Explanation
C) \(\frac{\sqrt{2}}{2}\)
To find the value of \(\sin \left(\frac{3 \pi}{4}\right)\), we can use the unit circle or the special angles in trigonometry.
In the unit circle, the angle \(\frac{3 \pi}{4}\) corresponds to the point \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) in the third quadrant.
The \(y\)-coordinate of this point gives the value of \(\sin \left(\frac{3 \pi}{4}\right)\), which is \(\frac{\sqrt{2}}{2}\).
Question Hard
Which expression is equivalent to sin 50° ?
A) cos 50°
B) cos 40°
C) tan 50°
D) sin 40°
▶️Answer/Explanation
B) cos 40°
To find the expression equivalent to \(\sin 50^\circ\), we can use the co-function identity for sine and cosine.
The co-function identity states:
\[
\sin (90^\circ – \theta) = \cos \theta
\]
For \(\theta = 50^\circ\):
\[
\sin 50^\circ = \cos (90^\circ – 50^\circ) = \cos 40^\circ
\]
Thus, the expression equivalent to \(\sin 50^\circ\) is:
\[ \boxed{\cos 40^\circ}~B \]
The figure shows the mast of a boat that is installed perpendicular to the deck of the boat. The mast is secured by a rope that is anchored to the deck. The rope measures 17 feet long and makes an angle of \(x^{\circ}\) with the mast. The point where the rope is attached to the mast is 15 feet above the deck. What is the value of \(\tan (x^{\circ})\) ?
A) \(\frac{5}{12}\)
B) \(\frac{8}{15}\)
C) \(\frac{9}{17}\)
D) \(\frac{12}{17}\)
▶️ Answer/Explanation
Answer: \(\frac{8}{15}\) or 0.533
In \(\triangle ABC\), using Pythagoras theorem: \(BC^2 = 17^2 – 15^2 = 289 – 225 = 64\), so \(BC = 8\). Then, \(\tan x = \frac{BC}{AB} = \frac{8}{15}\).