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Vector Algebra

Question
CBSEENMA12034147
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Prove that an angle inscribed in a semi-circle is a right angle.

Solution
Let O be the centre of the semi circle and AB be the diameter. Let P be any point on the circumference of the semi circle.
Let OA with rightwards arrow on top space equals space straight a with rightwards arrow on top comma space space then space OB with rightwards arrow on top space equals negative straight a with rightwards arrow on top.
Let OP with rightwards arrow on top space equals space straight r with rightwards arrow on top
therefore space space space AP with rightwards arrow on top space equals OP with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space straight r with rightwards arrow on top space minus space straight a with rightwards arrow on top space space space space space space space BP with rightwards arrow on top space equals space OP with rightwards arrow on top space minus space OB with rightwards arrow on top space equals straight r with rightwards arrow on top space minus space left parenthesis straight a with rightwards arrow on top right parenthesis space equals space straight r with rightwards arrow on top plus straight a with rightwards arrow on top AP with rightwards arrow on top space BP with rightwards arrow on top space equals space left parenthesis straight r with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis. space left parenthesis straight r with rightwards arrow on top space plus space straight a with rightwards arrow on top right parenthesis space equals space straight r with rightwards arrow on top squared space minus space straight a with rightwards arrow on top squared space space space space space space space space space space space space space equals straight a squared minus straight a squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because straight r space equals space straight a space as space OP space equals space OA close square brackets therefore space space space AP with rightwards arrow on top space space is space perpendicular space to space BP with rightwards arrow on top rightwards double arrow space space angle APB space space equals space 90 degree Hence space the space result. space space space space space space space space

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