Using Vectors, prove that angle in a semicircle is a right angle

Advertisement Remove all ads

#### Solution

Let sec AB be a diameter of a circle with centre C and P be any point on the circle other than A and B. Then

∠APB is an angle subtended on a semicircle

Let `bar(AC) = bar(CB) = bar(a) and bar(CP) = bar(r)`

Then `|bar(a)| = |bar(r)|` .... (1)

`bar(AP) = bar(AC) + bar(CP) = bar(a) + bar(r)= bar(r) + bar(a)`

`bar(BP) = bar(BC) + bar(CP) = -bar(CB) + bar(CP) = -bar(a) + bar(r)`

`:. bar(AP).bar(BP) = (bar(r) + bar(a)).(bar(r) - bar(a))`

`= bar(r).bar(r) - bar(r).bar(a) + bar(a).bar(r) - bar(a).bar(a)`

`= |bar(r)|^2 - |bar(a)|^2 = 0` ......`(∵ bar(r).bar(a) = bar(a) . bar(r))`

`:. bar(AP) ⊥ bar(BP)` ∴ ∠APB is a right angle

Hence, the angle subtended on a semicircle is the right angle.

Concept: Basic Concepts of Vector Algebra

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads