Question
In the given figure, $A D=A B=A C, B D$ is parallel to $C A$ and angle $\text{ACB}$
$=65^{\circ}$. Find $\angle DAC.$
$=65^{\circ}$. Find $\angle DAC.$
✓
Answer
We can see that the $\triangle ABC$ is an isosceles triangle with Side $AB =$ Side $AC.$
$\Rightarrow \angle ACB = \angle ABC$
As $\angle ACB = 65^\circ $
hence $\angle ABC = 65^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ACB + \angle CAB + \angle ABC = 180^\circ $
$65^\circ + 65^\circ + \angle CAB = 180^\circ $
$\angle CAB = 180^\circ − 130^\circ $
$\angle CAB = 50^\circ $ As $BD$ is parallel to $CA$
Therefore, $\angle CAB = \angle DBA$ since they are alternate angles.
$\angle CAB = \angle DBA = 50^\circ $
We see that $\triangle ADB$ is an isosceles triangle with Side $AD =$ Side $AB.$
$\Rightarrow \angle ADB = \angle DBA = 50^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ADB + \angle DAB + \angle DBA = 180^\circ $
$50^\circ + \angle DAB + 50^\circ = 180^\circ $
$\angle DAB = 180^\circ − 100^\circ = 80^\circ $
$\angle DAB = 80^\circ $
The $\angle DAC$ is the sum of $\angle DAB$ and $\text{CAB}.$
$\angle DAC = \angle CAB + \angle DAB$
$\angle DAC = 50^\circ + 80^\circ $
$\angle DAC = 130^\circ $
$\Rightarrow \angle ACB = \angle ABC$
As $\angle ACB = 65^\circ $
hence $\angle ABC = 65^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ACB + \angle CAB + \angle ABC = 180^\circ $
$65^\circ + 65^\circ + \angle CAB = 180^\circ $
$\angle CAB = 180^\circ − 130^\circ $
$\angle CAB = 50^\circ $ As $BD$ is parallel to $CA$
Therefore, $\angle CAB = \angle DBA$ since they are alternate angles.
$\angle CAB = \angle DBA = 50^\circ $
We see that $\triangle ADB$ is an isosceles triangle with Side $AD =$ Side $AB.$
$\Rightarrow \angle ADB = \angle DBA = 50^\circ $
Sum of all the angles of a triangle is $180^\circ $
$\angle ADB + \angle DAB + \angle DBA = 180^\circ $
$50^\circ + \angle DAB + 50^\circ = 180^\circ $
$\angle DAB = 180^\circ − 100^\circ = 80^\circ $
$\angle DAB = 80^\circ $
The $\angle DAC$ is the sum of $\angle DAB$ and $\text{CAB}.$
$\angle DAC = \angle CAB + \angle DAB$
$\angle DAC = 50^\circ + 80^\circ $
$\angle DAC = 130^\circ $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Prove that the line segment joining the mid$-$points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.
→The given figure shows a parallelogram $\text{ABCD}$ with area $324 sq. \ cm. P$ is a point in $AB$ such that $AP: PB = 1:2$.Find The area of $\triangle APD.$
→The difference between selling an article at $8\%$ profit and at $12\%$ profit is $Rs.72$. Find the cost price of the article and also the two selling prices.
→Find graphically the vertices of the triangle, whose sides are given by $3y = x + 18, x + 7y = 22$ and $y + 3x = 26.$
→In the given figure, ABCD is an isosceles trapezium in which $\angle CDA =2 x$ and $\angle BAD =3 x$.
Find all the angles of the trapezium.
→Find all the angles of the trapezium.
Construct a quadrilateral $ABCD$ in which $AB = 4.8\ cm, AC = 5.8\ cm, AD = 3.6\ cm, \angle A = 105^\circ $ and $\angle B = 60^\circ $
→A square brass plate of side x cm is 1 mm thick and weighs 5.44 kg. If 1 $cm ^3$ of brass weighs 8.5 g, find the value of x.
→$O$ is any point inside a rectangle $\text{ABCD}$.Prove that: $OB^2+ OD^2= OC^2+ OA^2$.
→A rectangular tank $30 \ cm \times 20 \ cm \times 12 \ cm$ contains water to a depth of $6 \ cm$ . A metal cube of side $10 \ cm$ is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in liters, that must be poured in the tank so that the metal cube is just submerged in the water.
→Two mobiles $S_1$ and $S_2$ are sold for $Rs. 10,490$ making $4\%$ profit on $S_1$ and $6\%$ on $S_2$. If the two mobiles are sold for $Rs.10,510,$ a profit of $6\%$ is made on $S_1$ and $4\%$ on $S_2$. Find the cost price of both the mobiles.
→