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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 $\angle OAB$.

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:

O D C A B

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


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