In the figure given below, find RS and PS using the information given in \(\triangle\)PSR.
Step 1: Understanding the Concept:
This problem involves solving a right-angled triangle using trigonometric ratios or the properties of a 30-60-90 triangle. Given one side and one angle, we can find the lengths of the other two sides.
Step 2: Key Formula or Approach:
We can use two methods:
Method 1: Trigonometric Ratios
\[\begin{array}{rl} \bullet & \text{\( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \)} \\ \bullet & \text{\( \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)} \\ \end{array}\]
Method 2: 30-60-90 Triangle Theorem
The sides of a 30-60-90 triangle are in the ratio \( 1 : \sqrt{3} : 2 \).
\[\begin{array}{rl} \bullet & \text{The side opposite 30\(^\circ\) is half the hypotenuse.} \\ \bullet & \text{The side opposite 60\(^\circ\) is \( \sqrt{3} \) times the side opposite 30\(^\circ\).} \\ \end{array}\]
Step 3: Detailed Explanation:
Given:
\[\begin{array}{rl} \bullet & \text{In \(\triangle\)PSR, \(\angle\)S = 90\(^\circ\).} \\ \bullet & \text{Hypotenuse PR = 12.} \\ \bullet & \text{\(\angle\)P = 30\(^\circ\).} \\ \bullet & \text{Therefore, \(\angle\)R = 180\(^\circ\) - 90\(^\circ\) - 30\(^\circ\) = 60\(^\circ\).} \\ \end{array}\]
Using Method 2 (30-60-90 Triangle Theorem):
1. Find RS: RS is the side opposite the 30\(^\circ\) angle (\(\angle\)P).
According to the theorem, the side opposite the 30\(^\circ\) angle is half the hypotenuse.
\[ RS = \frac{1}{2} \times PR = \frac{1}{2} \times 12 = 6 \]
2. Find PS: PS is the side opposite the 60\(^\circ\) angle (\(\angle\)R).
According to the theorem, the side opposite the 60\(^\circ\) angle is \( \sqrt{3} \) times the side opposite the 30\(^\circ\) angle.
\[ PS = \sqrt{3} \times RS = \sqrt{3} \times 6 = 6\sqrt{3} \]
Step 4: Final Answer:
The length of RS is 6 and the length of PS is 6\(\sqrt{3}\).
An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60\(^\circ\). To find the height of tree complete the activity. (\(\sqrt{3} = 1.73\))
Activity :
In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
Angle of elevation (\(\theta\)) = \(\angle\)BCA = 60\(^\circ\)
tan \(\theta\) = \(\frac{\boxed{\phantom{AB}}}{BC}\) \(\dots\) (I)
tan 60\(^\circ\) = \(\boxed{\phantom{\sqrt{3}}}\) \(\dots\) (II)
\(\frac{AB}{BC} = \sqrt{3}\) \(\dots\) (From (I) and (II))
AB = BC \(\times\) \(\sqrt{3}\) = 10\(\sqrt{3}\)
AB = 10 \(\times\) 1.73 = \(\boxed{\phantom{17.3}}\)
\(\therefore\) height of the tree is \(\boxed{\phantom{17.3}}\) m.
If \( \cos^2(10^\circ) \cos(20^\circ) \cos(40^\circ) \cos(50^\circ) \cos(70^\circ) = \alpha + \frac{\sqrt{3}}{16} \cos(10^\circ) \), then \( 3\alpha^{-1} \) is equal to:
सरस्वती विद्यालय, कोल्हापुर में मनाए गए 'शिक्षक दिवस' समारोह का 70 से 80 शब्दों में वृत्तांत लेखन कीजिए।
(वृत्तांत में स्थल, काल, घटना का उल्लेख होना अनिवार्य है)
निम्नलिखित जानकारी के आधार पर 50 से 60 शब्दों में विज्ञापन तैयार कीजिए :