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
Question Hard
In right triangle \(A B C\), angle \(C\) is a right angle and \(\sin A=0.70\). What is the value of \(\cos B\) ?
▶️Answer/Explanation
Ans: 7/10
In right triangle \(ABC\) with angle \(C\) as the right angle, \(\sin A = 0.70\). We need to find the value of \(\cos B\).
Since \(\angle A\) and \(\angle B\) are complementary angles (adding up to \(90^\circ\)), we know that:
\[
\cos B = \sin A
\]
Given \(\sin A = 0.70\), we have:
\[
\cos B = 0.70
\]
So, the value of \(\cos B\) is:
\[
\boxed{0.70}
\]
Question Hard
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 \(\cos x^{\circ }-\sin y^{\circ }\)?
▶️Answer/Explanation
Ans: 0
\(x^{\circ }+y^{\circ }=90^{\circ }\)
\(x^{\circ }=90-y^{\circ }\)
Taking cosine both side ,
\[
\cos \left(x^{\circ}\right)=\cos \left(90-y^{\circ}\right)
\]
Using the cosine of the difference formula, \(\cos (a-b)=\cos (a) \cos (b)+\sin (a) \sin (b)\), we have:
\[
\cos \left(x^{\circ}\right)=\cos (90) \cos \left(y^{\circ}\right)+\sin (90) \sin \left(y^{\circ}\right)
\]
Since \(\cos (90)=0\) and \(\sin (90)=1\), the equation simplifies to:
\[
\begin{aligned}
& \cos \left(x^{\circ}\right)=0 \cdot \cos \left(y^{\circ}\right)+1 \cdot \sin \left(y^{\circ}\right) \\
& \cos \left(x^{\circ}\right)=\sin \left(y^{\circ}\right)\\
&\cos \left(x^{\circ}\right)-\sin \left(y^{\circ}\right)=0
\end{aligned}
\]
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:
\[ \boxed{\cos 40^\circ}~B \]
Question Hard
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 \(\tan \left(x^{\circ}\right)\) ?
▶️Answer/Explanation
$.533, 8 / 15$
In \( \triangle A B C \), Using Pythagoras theorem
\[
\begin{aligned}
A B^2+B C^2 & =A C^2 \\
B C^2 & =(17)^2-(15)^2 \\
& =\sqrt{289-225} \\
& =\sqrt{64} \\
B C & =8 \text { feet }
\end{aligned}
\]
Now, $\text{tan x}=\frac{BC}{AB}$
$\text{tan x}=\frac{8}{15}$