Question

Given that m IHJ = (3a + 6)° and m KHT = (5a - 18)°,
identify m IHJ.

Answer 1

Answer:

$m\angle {\rm IHJ} = 42^{\circ}$.

Step-by-step explanation:

Start by showing that $\triangle {\rm IHJ} \cong \triangle {\rm KHJ}$ by ${\rm SSS}$.

(Triangle $\triangle {\rm IHJ}$ and $\triangle {\rm KHJ}$ are congruent by side, side, side- all three pairs of corresponding sides are equal in length.)

From the diagram, $\triangle {\rm IHJ}$ and $\triangle {\rm KHJ}$ already share two pairs of sides of the same length:

• ${\rm IJ} = {\rm KJ}$ as denoted on the diagram by the short slash.
• ${\rm HJ} = {\rm HJ}$ (this side is shared between the two triangles.)

It is also given that $\angle {\rm HIJ} = 90^{\circ}$ and $\angle {\rm HKJ} = 90^{\circ}$.

However, $\angle {\rm HIJ} = \angle {\rm HKJ}$ combined with ${\rm IJ} = {\rm KJ}$ and ${\rm HJ} = {\rm HJ}$ would not be sufficient for showing that $\triangle {\rm IHJ} \cong \triangle {\rm KHJ}$ by side, angle, side (${\rm SAS}$.) Two triangles are congruent by ${\rm SAS}\!$ only if angle that is equal is between the two pairs of sides.

Since $\triangle {\rm IHJ}$ and $\triangle {\rm KHJ}$ are both right triangles, the Pythagorean Theorem would apply:

$\displaystyle {\rm HI} = \sqrt{({\rm HJ})^{2} - ({\rm IJ})^{2}}$ in $\triangle {\rm IHJ}$.

Likewise:

$\displaystyle {\rm HK} = \sqrt{({\rm HJ})^{2} - ({\rm KJ})^{2}}$ in $\triangle {\rm KHJ}$.

Since ${\rm IJ} = {\rm KJ}$:

$\begin{aligned}{\rm HI} &= \sqrt{({\rm HJ})^{2} - ({\rm IJ})^{2}} \\ &= \sqrt{({\rm HJ})^{2} - ({\rm KJ})^{2}} = {\rm HK} \end{aligned}$.

Therefore, $\triangle {\rm IHJ} \cong \triangle {\rm KHJ}$ by ${\rm SSS}$ (${\rm IJ} = {\rm KJ}$, ${\rm HJ} = {\rm HJ}$, and ${\rm HI} = {\rm HK}$.)

By congruency, $m\angle {\rm IHJ} = m\angle {\rm KHJ}$.

Thus, $3\, a + 6 = 5\, a - 18$. Solve this equation to get $a = 12$.

Therefore, $m\angle {\rm IHJ} = (3\, a + 6)^{\circ} = 42^{\circ}$.