Question

In $△ A B C$ , side $A B$ is produced to $D$ such that $B D = B C$ . If $∠ A = 70^{o}$ and $∠ B = 60^{o}$ , prove that $A D > A C$ .

Accepted Answer

Given,

In $△ A B C$ ,

$∠ A = 70^{o}$ and $∠ B = 60^{o}$
Based on the sum property of the triangle,
$∠ A + ∠ B + ∠ C = 180^{o}$
$\Rightarrow ∠ C = 180^{o} - ∠ A - ∠ B$
$\Rightarrow ∠ C = 180^{o} - 70^{o} - 60^{o}$
$∴ ∠ C = 50^{o}$

Consider $△ B C D$ ,
We know that $∠ C B D$ is the exterior angle of $∠ A B C$ .
So we get,
$∠ C B D = ∠ D A C + ∠ A C B$
By substituting the values in the above equation
$∠ C B D = 70^{o} + 50^{o}$
$\Rightarrow ∠ C B D = 120^{o}$

It is given that $B C = B D$
So we can write it as $∠ B C D = ∠ B D C$ .
Based on the sum property of the triangle
$∠ B C D + ∠ B D C + ∠ C B D = 180^{o}$
$\Rightarrow ∠ B C D + ∠ B D C = 180^{o} - ∠ C B D$
By substituting values in the above equation
$\Rightarrow ∠ B C D + ∠ B D C = 180^{o} - 120^{o}$
$\Rightarrow ∠ B C D + ∠ B D C = 60^{o}$
It can be written as
$2 ∠ B C D = 60^{o}$
Or, $∠ B C D = ∠ B D C = 30^{o}$

In $△ A C D$
It is given that $∠ A = 70^{o}$ and $∠ B = 60^{o}$
We can write it as
$∠ A C D = ∠ A C B + ∠ B C D$
By substituting the values we get,
$\Rightarrow ∠ A C D = 50^{o} + 30^{o}$
$\Rightarrow ∠ A C D = 80^{o}$

So we get to know that $∠ A C D$ is the greatest angle and the side opposite to it, i.e. $A D$ is the longest side.

Therefore, it is proved that $A D > A C$ .

R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: EXERCISE 9B

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: EXERCISE 9B

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: EXERCISE 9B with Hints & Solutions

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