Interactive multiple-choice quiz: اختبار الوحدة الرابعة: المثلثات القائمة وحساب المثلثات - ريفيل
الرياضيات ريفيل الصف العاشر المتقدم الفصل الدراسي الأول للعام 2026 - 2027 مراجعة شاملة لدروس الوحدة الرابعة: (Right Triangles and Trigonometry) إعداد الأستاذ: Mr. Aghead
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اختبار شهادة تدريبي مؤقت للصف والمادة والفصل نفسه.
اختر إجابة واحدة لكل سؤال. عند الاختيار ستظهر النتيجة فورًا: الأخضر صحيح، والأحمر خطأ، وسيظهر تفسير الإجابة مباشرة إن كان متوفرًا. وبعد آخر سؤال ستظهر الدرجة النهائية تلقائيًا.
Find the geometric mean between 12 and 15. Write your answer in simplest radical form, if necessary.
Explanation
The geometric mean of a and b is $\sqrt{a \cdot b}$. Here, $\sqrt{12 \cdot 15} = \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}$.
Question 2
2
Points: 1
Write a similarity statement identifying the three similar right triangles in the figure.
Explanation
Based on the Right Triangle Similarity Theorem, the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other, matching corresponding vertices.
Question 3
3
Points: 1
Use a Pythagorean triple to find the value of x.
Explanation
Using the Pythagorean triple (5, 12, 13) multiplied by 4: 5(4)=20, 12(4)=48, so 13(4)=52.
Question 4
4
Points: 1
The bottom end of a ramp at a warehouse is 10 feet from the base of the main dock, and the ramp is 11 feet long. How high is the dock?
Explanation
Using the Pythagorean theorem: $h = \sqrt{11^2 - 10^2} = \sqrt{121 - 100} = \sqrt{21} \approx 4.58$. Rounded to the nearest tenth, it is 4.6 ft.
Question 5
5
Points: 1
Determine the coordinates of the midpoint M of $\overline{AB}$ with endpoints A(-7, 9, 4) and B(5, -3, -4).
Explanation
The midpoint formula in 3D is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})$. Calculating: $(\frac{-7+5}{2}, \frac{9-3}{2}, \frac{4-4}{2}) = (-1, 3, 0)$.
Question 6
6
Points: 1
Graph a rectangular solid that contains the given point and the origin as vertices. Label the coordinates of the vertex R(3, -4, 1).
Explanation
The graph in C correctly represents the point (3, -4, 1) in a 3D coordinate system relative to the origin.
Question 7
7
Points: 1
Find the values of x and y.
Explanation
In a $30^\circ-60^\circ-90^\circ$ triangle, if the side opposite $30^\circ$ is 15, then the hypotenuse y = 2 × 15 = 30, and the side opposite $60^\circ$ is $x = 15\sqrt{3}$.
Question 8
9
Points: 1
Find $m\angle Z$ to the nearest tenth.
Explanation
In the \right triangle, $\cos Z = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{10}{14}$. Thus, $Z = \arccos(\frac{10}{14}) \approx 44.4^\circ$.
Question 9
10
Points: 1
Use a special \right triangle to express the trigonometric ratio as a fraction and as a decimal to the nearest hundredth if necessary: $\tan 30^\circ$.
Explanation
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$, which is approximately 0.577, rounded to 0.58.
Question 10
11
Points: 1
Solve the right triangle. Round side measures to the nearest tenth and angle measures to the nearest degree.
Braylen stands on an 8-foot platform and sights a swimmer at an angle of depression of $5^\circ$. If Braylen is 6 feet tall, how far away is the swimmer from the base of the platform to the nearest foot?
Explanation
Total height is 8 + 6 = 14 feet. The distance d is calculated by $\tan 5^\circ = \frac{14}{d}$, so $d = \frac{14}{\tan 5^\circ} \approx 160.02$ feet.
Question 12
13
Points: 1
Use trigonometry to find the area of $\Delta JKL$ to the nearest tenth.
Explanation
Area $= \frac{1}{2}bc \sin A = \frac{1}{2}(6)(10.5)\sin 83^\circ \approx 31.5 \times 0.9925 \approx 31.3$.
Question 13
14
Points: 1
A student can see a water tower from the edge of a soccer field. The edge of the field is about 110 feet from the water tower, and the tower stands at a height of 32.5 feet. What is the angle of elevation if the eye level of the student viewing the tower is 6 feet above the ground? Round to the nearest tenth.
Explanation
The relative height is 32.5 - 6 = 26.5 feet. $\tan \theta = \frac{26.5}{110} \Rightarrow \theta = \arctan(0.2409) \approx 13.5^\circ$.
Question 14
16
Points: 1
Find the value of d in the figure.
Explanation
Using the Law of Sines: The third angle is $180 - (68 + 31) = 81^\circ$. $\frac{d}{\sin 31^\circ} = \frac{100}{\sin 81^\circ} \Rightarrow d = \frac{100 \sin 31^\circ}{\sin 81^\circ} \approx 52.1$ ft.
Question 15
17
Points: 1
Determine whether the triangle has no solution, one solution, or two solutions. Then, solve the triangle: $m\angle A = 133^\circ, a = 9, b = 7$.
Explanation
Since A is obtuse and a > b, there is exactly one solution. Using Law of Sines: $\sin B = \frac{7 \sin 133^\circ}{9} \approx 0.568 \Rightarrow B \approx 35^\circ$. $C = 180 - 133 - 35 = 12^\circ$. $c = \frac{9 \sin 12^\circ}{\sin 133^\circ} \approx 2.6$.
Question 16
18
Points: 1
Find the value of x to the nearest degree.
Explanation
Using the Law of Cosines: 22 = 1.12 + 1.52 - 2(1.1)(1.5)cos x. $4 = 1.21 + 2.25 - 3.3\cos x \Rightarrow 0.54 = -3.3\cos x \Rightarrow \cos x \approx -0.1636 \Rightarrow x \approx 99^\circ$.
Question 17
19
Points: 1
Solve $\Delta ABC$ when b = 10.2, c = 9.3, and $m\angle A = 26^\circ$. Find a.
Explanation
Using the Law of Cosines: $a^2 = 10.2^2 + 9.3^2 - 2(10.2)(9.3)\cos 26^\circ \approx 104.04 + 86.49 - 189.72(0.8988) \approx 20.0 \Rightarrow a \approx 4.5$.
Question 18
20
Points: 1
Two radar stations 2.4 miles apart are tracking an airplane. The distance between Station A and the plane is 7.4 miles. The distance between Station B and the plane is 6.9 miles. What is the angle of elevation from Station A to the plane?
Explanation
Using the Law of Cosines for the triangle with sides 2.4, 7.4, and 6.9: $6.9^2 = 2.4^2 + 7.4^2 - 2(2.4)(7.4)\cos A \Rightarrow 47.61 = 5.76 + 54.76 - 35.52\cos A \Rightarrow \cos A \approx 0.363 \Rightarrow A \approx 69^\circ$.
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