Step 1: Understanding the Concept:
This question is based on the Inscribed Angle Theorem in circle geometry. The theorem states that the measure of an angle inscribed in a circle is half the measure of its intercepted arc.
Step 2: Key Formula or Approach:
The formula derived from the Inscribed Angle Theorem is:
\[ \text{Measure of Inscribed Angle} = \frac{1}{2} \times \text{Measure of Intercepted Arc} \]
Or, rearranging for the arc:
\[ \text{Measure of Intercepted Arc} = 2 \times \text{Measure of Inscribed Angle} \]
Step 3: Detailed Explanation:
We are given the measure of the inscribed angle, \(\angle\)ACB.
\[ \angle\text{ACB} = 65^{\circ} \]
The intercepted arc is AXB.
Using the formula from Step 2:
\[ \text{Measure of arc AXB} = 2 \times \angle\text{ACB} \]
\[ \text{Measure of arc AXB} = 2 \times 65^{\circ} \]
\[ \text{Measure of arc AXB} = 130^{\circ} \]
Step 4: Final Answer:
The measure of the intercepted arc AXB is 130\(^{\circ}\).
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB.
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 :