Question

In the given figure, if $$AB=AC$$, $CH=CB$ and $HK||BC$, then the value of $\mathrm{\angle }HCK$ is

A) ${30}^{\circ }$
B) ${35}^{\circ }$
C) ${40}^{\circ }$
D) ${45}^{\circ }$

Answer 1

Hint: First find the value of the angle $\mathrm{\angle }BAC$ using the fact that of supplementary angles and then use the property that if the sides of the triangle are same then their corresponding angles are also the same. Then use the given data that $HK||BC$, to approach the desired result.

Complete step by step answer:
We have given that $$AB=AC$$, $CH=CB$, exterior angle $\mathrm{\angle }A={140}^{\circ }$ and $HK||BC$ in the adjoint figure.
The goal of the problem is to find the measure of the angle $\mathrm{\angle }HCK$.
As we have an exterior angle $\mathrm{\angle }A={140}^{\circ }$, and $\mathrm{\angle }BAC$ and $\mathrm{\angle }A$ are supplementary angles. So, we can write
$\mathrm{\angle }BAC=180-\mathrm{\angle }A$
$\mathrm{\angle }BAC={180}^{\circ }-{140}^{\circ }$
$\mathrm{\angle }BAC={40}^{\circ }$
Now, we have given that$AB=AC$, it means that the given triangle ABC is an isosceles trinagle so their base angles are same, therefore
$\mathrm{\angle }ACB=\mathrm{\angle }ABC$
In a triangle$\mathrm{\Delta }ABC$, we have
$\mathrm{\angle }BAC={40}^{\circ }$and $\mathrm{\angle }ACB=\mathrm{\angle }ABC$
We know that the sum of the angles in a triangle is ${180}^{\circ }$then we can write
$\mathrm{\angle }ABC+\mathrm{\angle }ACB+\mathrm{\angle }BAC={180}^{\circ }$
Substitute the value of the angles $\mathrm{\angle }BAC={40}^{\circ }$and $\mathrm{\angle }ACB=\mathrm{\angle }ABC$ then we have
$\mathrm{\angle }ABC+\mathrm{\angle }ABC+{40}^{\circ }={180}^{\circ }$
$\Rightarrow 2\mathrm{\angle }ABC={180}^{\circ }-{40}^{\circ }$
$\Rightarrow \mathrm{\angle }ABC={\displaystyle \frac{{140}^{\circ }}{2}}$
$\Rightarrow \mathrm{\angle }ABC={70}^{\circ }$
Then we also have:
$\mathrm{\angle }ACB=\mathrm{\angle }ABC={70}^{\circ }$
It is also given that the transversal lines $HK$ is parallel to the $BC$, then using the property of transversal lines, we have:
$\mathrm{\angle }ACB=\mathrm{\angle }ABC=\mathrm{\angle }AHK=\mathrm{\angle }AKH$
$\Rightarrow \mathrm{\angle }AHK=\mathrm{\angle }AKH={70}^{\circ }$
We can see that $\mathrm{\angle }AKH$ and $\mathrm{\angle }HKC$ are supplementary angles, so we have
$\mathrm{\angle }AKH+\mathrm{\angle }HKC=180$
Substituting the values $\mathrm{\angle }AKH={70}^{\circ }$, then
${70}^{\circ }+\mathrm{\angle }HKC=180$
$\mathrm{\angle }HKC={180}^{\circ }-{70}^{\circ }$
$\mathrm{\angle }HKC={110}^{\circ }$
We also have given that $CH=CB$, then the corresponding angles are also same, that is
$\mathrm{\angle }HBC=\mathrm{\angle }BHC$
The angle $\mathrm{\angle }HBC$ is equal to the angle $\mathrm{\angle }ABC$. So, we can write as
$\Rightarrow \mathrm{\angle }HBC=\mathrm{\angle }BHC={70}^{\circ }$
We can see in the figure that,
$\Rightarrow \mathrm{\angle }AHK+\mathrm{\angle }KHC+\mathrm{\angle }BHC={180}^{\circ }$
Substitute the values $\mathrm{\angle }AHK={70}^{\circ }$ and $\mathrm{\angle }BHC={70}^{\circ }$ into the equation
${70}^{\circ }+\mathrm{\angle }KHC+{70}^{\circ }={180}^{\circ }$
$\Rightarrow \mathrm{\angle }KHC={180}^{\circ }-{140}^{\circ }$
$\Rightarrow \mathrm{\angle }KHC={40}^{\circ }$
Now, in the triangles $\mathrm{\Delta }KHC$, we have
$$\Rightarrow \mathrm{\angle }KHC={40}^{\circ }$$ and $\mathrm{\angle }HKC={110}^{\circ }$
We know that the sum of the angles of the triangle is ${180}^{\circ }$, so we can write
$$\Rightarrow \mathrm{\angle }KHC+\mathrm{\angle }HKC+\mathrm{\angle }HCK={180}^{\circ }$$
Substituting the values $$\mathrm{\angle }KHC={40}^{\circ }$$ and $\mathrm{\angle }HKC={110}^{\circ }$ into the equation,
$$\Rightarrow {40}^{\circ }+{110}^{\circ }+\mathrm{\angle }HCK={180}^{\circ }$$
$$\Rightarrow \mathrm{\angle }HCK={180}^{\circ }-{150}^{\circ }$$
$$\Rightarrow \mathrm{\angle }HCK={30}^{\circ }$$

Therefore, the required measurement of the angle has the value $$\mathrm{\angle }HCK={30}^{\circ }$$.

Note:
If the sum of two angles is ${180}^{\circ }$then these angles are said as supplementary angles and of the sum of two angles is ${90}^{\circ }$, then these angles are called complementary angles.