Question

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In figure AOB is a line and ray OC is perpendicular to AB . If OD is another ray lying between OB and OC . Prove that angle cod =1/2 (angle AOD- angle BOD)

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Answer 1

Answer:

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Step-by-step explanation:

COB = 90°

AOC = 90°

COD = AOC - AOD

= 90 ° - AOD

So,

COD + AOD = 90°

Since, OC is perpendicular to AB

Therefore, AOD = 2/3 AOC

= 150/3

= 50°

So,

COD = 90° - 50°

= 40°

BOD - AOD = 80°

So,

1/2 × 80° = 40°

LHS = RHS

Hence proved

Answer 2

Correct Question :-

In figure AOB is a line and ray OC is perpendicular to AB . If OD is another ray lying between OB and OC . Prove that angle cod =1/2 (angle BOD- angle AOD)

Given:-

• In figure AOB is a line and ray OC is perpendicular to AB.
• OD is another ray lying between OB and OC .

To prove:-

$\sf \angle {COD}=\dfrac {1}{2}(\angle BOD-\angle AOD)$

Proof:-

Here ray CO is a Perpendicular to line AOB

${:}\longrightarrow\sf m\angle BOC=m\angle AOC=90°\dots (1)$

• And ,

$\sf \angle COD=\angle AOC-\angle AOD$

${:}\longrightarrow\sf \angle COD=90°-\angle AOD \left [\because from\: eq (1)\right]$

• Simplify

${:}\longrightarrow\sf \angle COD+\angle AOD=90°\dots(2)$

• Here,

CO is Perpendicular to AB $\sf \left[\because Given\right]$

${:}\longrightarrow\sf \angle AOD=\dfrac {2}{3}\angle AOC$

${:}\longrightarrow\sf \angle AOD=\dfrac {150}{3}$

${:}\longrightarrow\sf m\angle AOD=50°\dots (3)$

• Substitute the value in eq (1)

${:}\longrightarrow\sf \angle COD+50=90°$

${:}\longrightarrow\sf \angle COD=90°-50°$

${:}\longrightarrow\sf m\angle COD=40°\dots (4)$

• Again,

Here All angles are present on one line AB.Hence measure of the angles is 180°

${:}\longrightarrow\sf \angle AOD+\angle BOD=180°$

${:}\longrightarrow\sf 50°+\angle BOD=180°$

${:}\longrightarrow\sf \angle BOD=180-50$

${:}\longrightarrow\sf m\angle BOD=130°\dots (5)$

• From eq (3) and eq (5)

${:}\longrightarrow\sf \dfrac {1}{2}(\angle BOD-\angle AOD)$

${:}\longrightarrow\sf \dfrac {1}{2}(130-50)$

${:}\longrightarrow\sf \dfrac {80}{2}$

${:}\longrightarrow\sf 40°\dots (6)$

• Combine eq (4) and eq (6)

${:}\longrightarrow\sf m\angle COD=\dfrac {1}{2}(\angle BOD-\angle AOD)$

${:}\longrightarrow\sf 40°=40°$

$\therefore\underline{\underline{\sf Hence\: Proved.}}$