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प्रश्न
In a right-angled triangle ABC, ∠B = 90°, BD = 3, DC = 4, and AC = 13. A point D is inside the triangle such as ∠BDC = 90°.
Find the values of 3 - 2 cos ∠BAC + 3 cot ∠BCD
उत्तर
ΔBDC is a right-angled triangle.
∴ BC2
= BD2 +DC2
= 32 + 42
= 9 + 16
= 25
⇒ BC = 5cm
ΔABC is a right-angled triangle.
∴ AB2
= AC2 - BC2
= 132 - 52
= 169 - 25
= 144
⇒ AB = 12cm
3 - 2 cos ∠BAC + 3 cot ∠BCD
= `3 - 2 xx "AB"/"AC" + 3 xx "DC"/"BD"`
= `3 - 2 xx (12)/(13) + 3 xx (4)/(3)`
= `3 - (24)/(13) + 4`
= `7 - (24)/(13)`
= `(91 - 24)/(13)`
= `(67)/(13)`.
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