Question

Prove that the angle in a semi- circle is a right angle .​

Answer 1

Hi !!!

Solution :- Let O be the centre of semi- circle with AOB as its diameter.Let P be a point on the Circle, so that <APB is an angle in the semi-circle , join OP . Let O be taken as origin .

Let the position vectors of A , B and P be a vector , -a vector and r vector respectively.

Clearly, OA = OB = OP

Now, AP vector= (r-a) and BPvector = (r+a)

•°• AP * BP = (r-a)(r+a) = r²-a² = OP² -OA² =0

•°• AP perpendicular BP ,i.e <APB = 90°

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Answer 2

AnswEr:

$\sf\underline\pink{\:\:\:\:\:\:Given:-\:\:\:\:\:\:\:}$

PQ is a diameter of a circle C(O,r) and $\angle$ PRQ is an angle in semi-circle.

$\sf\underline\orange{\:\:\:\:\:\:To\:Prove:-\:\:\:\:\:\:\:}$

$\angle$ PRQ = 90°

$\sf\underline\blue{\:\:\:\:\:\:Proof:-\:\:\:\:\:\:\:}$

We know that the angle subtended by an arc of a circle at its centre is twice the angle formed by the same arc at a point on the circle. So, we have

$\\ \qquad \quad \sf \angle \: POQ = 2 \: \angle \: PRQ \\ \\ \\ \implies \sf \: 180 \degree \: = 2 \: \angle \: PRQ \qquad \: ( \because \: POQ \: is \: a \: straight \: line) \\ \\ \\ \implies \sf \angle \: PRQ = 90 \degree$