Find the best tutors and institutes for Class 10 Tuition
Q2:
In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

Solution :
Given:
- $\triangle ODC \sim \triangle OBA$
- $\angle BOC = 125^{\circ}$
- $\angle CDO = 70^{\circ}$
- $DB$ and $AC$ are straight lines intersecting at $O$.
To Find:
- $\angle DOC$
- $\angle DCO$
- $\angle OAB$
Visual Representation:
Step 1: Finding $\angle DOC$
Since $DB$ is a straight line, the angles $\angle DOC$ and $\angle BOC$ form a linear pair. [Axiom: The sum of angles on a straight line is $180^{\circ}$].
$\angle DOC + \angle BOC = 180^{\circ}$
$\angle DOC + 125^{\circ} = 180^{\circ}$
$\angle DOC = 180^{\circ} - 125^{\circ}$
$\angle DOC = 55^{\circ}$
Step 2: Finding $\angle DCO$
Consider $\triangle ODC$. The sum of the interior angles of a triangle is always $180^{\circ}$. [Theorem: Angle Sum Property of a Triangle].
$\angle ODC + \angle DOC + \angle DCO = 180^{\circ}$
Given $\angle ODC = 70^{\circ}$ and we found $\angle DOC = 55^{\circ}$.
$70^{\circ} + 55^{\circ} + \angle DCO = 180^{\circ}$
$125^{\circ} + \angle DCO = 180^{\circ}$
$\angle DCO = 180^{\circ} - 125^{\circ}$
$\angle DCO = 55^{\circ}$
Step 3: Finding $\angle OAB$
It is given that $\triangle ODC \sim \triangle OBA$. [Definition: If two triangles are similar, their corresponding angles are equal].
Therefore, $\angle OAB = \angle OCD$ (which is the same as $\angle DCO$).
Since $\angle DCO = 55^{\circ}$, it follows that:
$\angle OAB = 55^{\circ}$
Final Answer:
$\angle DOC = 55^{\circ}$, $\angle DCO = 55^{\circ}$, and $\angle OAB = 55^{\circ}$.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.3
- Q1(i): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (i)
- Q1(ii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (ii)
- Q1(iii): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iii)
- Q1(iv): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (iv)
- Q1(v): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (v)
- Q1(vi): State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : (vi)
- Q3: Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
- Q4: In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.
- Q5: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.
- Q6: In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.
- Q7(i): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (i) $\triangle AEP \sim \triangle CDP$
- Q7(ii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (ii) $\triangle ABD \sim \triangle CBE$
- Q7(iii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iii) $\triangle AEP \sim \triangle ADB$
- Q7(iv): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iv) $\triangle PDC \sim \triangle BEC$
- Q8: $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
- Q9(i): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC \sim \triangle AMP$
- Q9(ii): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (ii) $\frac{CA}{PA} = \frac{BC}{MP}$
- Q10(i): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (i) $\frac{CD}{GH} = \frac{AC}{FG}$
- Q10(ii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (ii) $\triangle DCB \sim \triangle HGE$
- Q10(iii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (iii) $\triangle DCA \sim \triangle HGF$
- Q11: In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.
- Q12: Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.
- Q13: $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
- Q14: Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
- Q15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
- Q16: If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
Top Tutors who teach Triangles
I am a teacher of maths and science, having a experience of 8+ years in teaching, done my schooling from CBSe board and degree from IP university.
For me, teaching is part of learning. As I am teaching I am also learning from my learners as they are diverse. Each learner in the classroom brings his or her unique background and ability. These are the elements that I invest in my teaching. Moreover, I allow learners to be themselves. This is a way of creating a conducive environment for learning.
Its been a really nice teaching experience with Mr.Amit so far. He is very knowledgeable and a patient teacher. He always clarifies the concepts clearly and ensures good understanding.
I am a teacher of Mathematics. I am teaching in school since 2013. I have a degree in Education. I have done Masters in Mathematics.
Hi, I'm Ravi Shankar, a Mathematics educator with 10+ years of teaching experience. I help Class 10 students strengthen their concepts, improve application skills, and prepare systematically for important examinations. At this stage, simply knowing formulas isn't enough. Students need to recognise the appropriate method, interpret questions correctly, solve efficiently and avoid losing marks through preventable mistakes. My classes focus on: • Concept mastery • Algebra and functions • Geometry • Trigonometry • Statistics and probability • Application-based problems • Multi-step questions • Past-paper practice • Error analysis • Time management and exam strategy I use regular assessments to identify recurring mistakes and areas requiring additional attention. My approach: Concept → Application → Problem Solving → Exam Practice → Analysis The goal is to help students approach examinations with greater confidence, accuracy, and consistency. Book an introductory session to understand where your child currently stands and what needs to improve.
I highly appreciate Sir’s dedication and clarity in teaching. His systematic approach helped me score confidently in board examinations.
I am a mathematics expert . I am teaching this subject since a decade . I ve handled thousands of students and bring numerous centum results in past . You will able to learn maths in the most pragmatic approach with me and develop a love for this subject .
Awesome humble best tutor for mathematics you and economic can just close your eyes and select for your class. You doesn't have to follow your kid he will take care of the student.
Find more Tutor for Triangles in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
Download free CBSE - Class 10 Mathematics Triangles EXERCISE 6.3 worksheets
Download Now