Question
Download Solution PDFत्रिभुज ABC में, कोण C = 90° है, AC = 2√3 सेमी और BC = 1 सेमी है। (13sin²A + 6sec²A + cosec²A) का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
त्रिभुज ABC, कोण C = 90°
AC = 2√3 सेमी
BC = 1 सेमी
प्रयुक्त सूत्र:
पाइथागोरस प्रमेय: AB² = AC² + BC²
sin A = सम्मुख भुजा / कर्ण = BC / AB
sec A = कर्ण / आसन्न भुजा = AB / AC
cosec A = कर्ण / सम्मुख भुजा = AB / BC
गणना:
पाइथागोरस प्रमेय का उपयोग करने पर, AB:
AB² = (2√3)² + 1²
AB² = 12 + 1 = 13
AB = √13 सेमी
अब,
sin A = BC / AB = 1 / √13
sec A = AB / AC = √13 / (2√3)
cosec A = AB / BC = √13 / 1 = √13
दिए गए व्यंजक में मान प्रतिस्थापित कीजिए:
13sin²A + 6sec²A + cosec²A
⇒ 13(1/√13)² + 6(√13 / (2√3))² + (√13)²
⇒ 13(1/13) + 6(13/12) + 13
⇒ 1 + 13/2 + 13
⇒ 14 + 6.5 = 20.5 = 41/2 = \(20 \frac{1}{2}\)
इसलिए, (13sin²A + 6sec²A + cosec²A) का मान \(20 \frac{1}{2}\) है।
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