Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q2 In Fig. 6.35, $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 \degree$ and $\angle CDO = 70 \degree$ . Find $\angle DOC , \angle DCO , \angle OAB$

Answers (1)

best_answer

Given : $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 \degree$ and $\angle CDO = 70 \degree$

$\angle DOC+\angle BOC=180 \degree$ (DOB is a straight line)

$\Rightarrow \angle DOC+125 \degree=180 \degree$

$\Rightarrow \angle DOC=180 \degree-125 \degree$

$\Rightarrow \angle DOC=55 \degree$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 \degree$

$\Rightarrow 55 \degree+ 70 \degree+\angle DCO=180 \degree$

$\Rightarrow \angle DCO+125 \degree=180 \degree$

$\Rightarrow \angle DCO=180 \degree-125 \degree$

$\Rightarrow \angle DCO=55 \degree$

Since , $\Delta ODC \sim \Delta OBA$ , so

$\Rightarrow\angle OAB= \angle DCO=55 \degree$ ( Corresponding angles are equal in similar triangles).

Posted by

seema garhwal

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE introduces remote exam system to evaluate school counsellors; 10,000 to be assessed in phases

anam-khan

CBSE introduces new assessment scheme for Class 10 students in Middle East

anam-khan

CBSE Class 10 Second Board Exams 2026: LOC schedule out; fees, guidelines for improvement exam

Latest Question