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SSC CHSL 15 Nov 2015 Shift 2 Paper

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Question : 125 of 200

Marks: +1, -0

In

\Delta ABC, AD \perp BC

AD^2 = BD \times DC

. Then measure of

\angle BAC

is:

75°
90°
45°
60°

Solution:

Given : In

\Delta ABC, AD \perp BC

and

AD^2 = BD \times DC

To find :

\angle BAC

Solution : In right

\Delta ADB

and

\Delta ADC,

if we apply Pythagoras Theorem,

⇒

(AB)^2 = (AD)^2 + (BD)^2

---(i)

and ⇒

(AC)^2 = (AD)^2 + (DC)^2

------(ii)

Adding equations (i) and (ii), we get :

⇒

(AB)^2 + (AC)^2 = 2(AD)^2

+ (BD)^2 + (DC)^2

⇒

(AB)^2 + (AC)^2 = 2(BD)(DC)

+ (BD)^2 + (DC)^2

[Given]

⇒

(AB)^2 + (AC)^2 = (BD + DC)^2

⇒

(AB)^2 + (AC)^2 = (BC)^2

Hence,

\Delta ABC

is a right triangle right angled at A.

∴ ∠BAC = 90°

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