Question

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

Answer 1

`m\angle {\rm IHJ} = 42^(\circ)`.

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)`.