Question

1. shira is given a 30°-60°-90° triangle with a hypotenuse that is 10 units long. she labels the other two sides as shown in the image. did she solve this triangle correctly? explain.

1. triangle abc is a right triangle. the measure of ∠a = 60° and ∠b is a right angle. which of the following statements is not true?

a) \\(\sin a = \frac{\sqrt{3}}{2}\\)
b) \\(\cos a = 0.5\\)
c) \\(\tan a = \sqrt{3}\\)
d) \\(\sin a = 0.5\\)

1. angle k is in standard position and the point (3,6) is on its terminal side.

a. draw the angle on the axes.
b. find the exact value of \\(\cos k\\).
c. find \\(m\angle k\\). round to the nearest tenth degree.

Explanation:

Question 12

Step1: Recall 30-60-90 triangle ratios

In a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest (let's call it \(x\)), the side opposite \(60^\circ\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Identify sides in Shira's triangle

Hypotenuse is 10, so \(2x = 10 \implies x = 5\). The side opposite \(30^\circ\) should be \(x = 5\), and the side opposite \(60^\circ\) should be \(x\sqrt{3}=5\sqrt{3}\). But in the triangle, the \(30^\circ\) angle is adjacent to the side labeled 5 and opposite to \(5\sqrt{3}\)? Wait, no—wait, the right angle: let's check the angles. The \(30^\circ\) angle: the side adjacent to \(30^\circ\) (if we consider \(30^\circ\) angle) should be the side opposite \(60^\circ\), and the side opposite \(30^\circ\) should be the shorter leg. Wait, in the diagram, the hypotenuse is 10, one leg is 5 (adjacent to \(30^\circ\)), and the other leg is \(5\sqrt{3}\) (opposite \(30^\circ\)? No, wait, the right angle: let's see the angles. The triangle has \(30^\circ\), \(60^\circ\), and right angle. So the side opposite \(30^\circ\) should be the shortest side. In Shira's triangle, the side labeled 5: is that opposite \(30^\circ\)? Wait, the \(30^\circ\) angle: the side opposite \(30^\circ\) should be the leg opposite, but in the diagram, the hypotenuse is 10, one leg is 5 (horizontal), one leg is \(5\sqrt{3}\) (vertical), and the \(30^\circ\) angle is at the left vertex. So the side opposite \(30^\circ\) is the vertical leg? Wait no, the \(30^\circ\) angle: the angle at the left has \(30^\circ\), so the side opposite to \(30^\circ\) is the vertical leg (length \(5\sqrt{3}\))? No, that's not right. Wait, no—wait, in a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, the side opposite \(30^\circ\) is the shortest leg. So if hypotenuse is 10, shortest leg (opposite \(30^\circ\)) is 5, and the longer leg (opposite \(60^\circ\)) is \(5\sqrt{3}\). Now, in the diagram, the \(30^\circ\) angle: the side adjacent to \(30^\circ\) is the longer leg (\(5\sqrt{3}\))? No, wait, no—let's label the triangle properly. Let's say the right angle is at the top right. Then the vertices: left vertex (30°), top right (right angle), bottom vertex (60°). So the sides: left to top right: 5 (horizontal), top right to bottom: \(5\sqrt{3}\) (vertical), left to bottom: 10 (hypotenuse). Now, the angle at left is 30°: the side opposite to 30° is the vertical leg (\(5\sqrt{3}\))? No, that's the side opposite 60° (since bottom angle is 60°). Wait, no—angle at bottom is 60°, so side opposite 60° is horizontal leg (5)? No, this is the mistake. Wait, the side opposite 30° should be the shortest leg. So hypotenuse 10, so shortest leg (opposite 30°) is 5, longer leg (opposite 60°) is \(5\sqrt{3}\). So in the triangle, the 30° angle: the side opposite 30° should be 5, but in the diagram, the side opposite 30° (if 30° is at left) is the vertical leg (\(5\sqrt{3}\))? No, that's incorrect. Wait, no—wait, the angle at left is 30°, so the side opposite to 30° is the side from top right to bottom (vertical leg), which is \(5\sqrt{3}\)? No, that's the side opposite 60° (since bottom angle is 60°). Wait, I think Shira mixed up the legs. The side opposite 30° should be 5, and the side opposite 60° should be \(5\sqrt{3}\). But in the diagram, the side adjacent to 30° is 5 (horizontal leg), and the side opposite 30° is \(5\sqrt{3}\) (vertical leg), which is the longer leg, but the side opposite 30° should be the shorter leg. Wait, no—wait, the hypotenuse is 10, so t…

Step1: Recall trigonometric ratios for \(60^\circ\)

For an angle \(A = 60^\circ\) in a right triangle:

• \(\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866\)
• \(\cos(60^\circ) = 0.5\)
• \(\tan(60^\circ) = \sqrt{3}\)

Step2: Evaluate each option

• Option A: \(\sin A = \frac{\sqrt{3}}{2}\) – True, since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).
• Option B: \(\cos A = 0.5\) – True, since \(\cos(60^\circ) = 0.5\).
• Option C: \(\tan A = \sqrt{3}\) – True, since \(\tan(60^\circ) = \sqrt{3}\).
• Option D: \(\sin A = 0.5\) – False, since \(\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866

eq 0.5\) (0.5 is \(\sin(30^\circ)\)).

Step1: Recall \(\cos K\) definition

\(\cos K = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r}\), where \(x = 3\), \(y = 6\), and \(r = \sqrt{x^2 + y^2}\) (distance from origin to \((3,6)\)).

Step2: Calculate \(r\)

\(r = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\).

Step3: Calculate \(\cos K\)

\(\cos K = \frac{x}{r} = \frac{3}{3\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\) (rationalizing the denominator).

Answer:

No, she did not solve it correctly. In a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, the side opposite \(30^\circ\) (shorter leg) should be \(5\) (since hypotenuse \(= 10 = 2x \implies x = 5\)), and the side opposite \(60^\circ\) (longer leg) should be \(5\sqrt{3}\). However, in her triangle, the side adjacent to \(30^\circ\) is \(5\) (which should be the longer leg, \(5\sqrt{3}\)) and the side opposite \(30^\circ\) is \(5\sqrt{3}\) (which should be the shorter leg, \(5\)), so the legs are mixed up.

Question 13