91Ó°ÊÓ

Label the given diagram as:

As$\stackrel{→}{OB}âŠ\yen l$,therefore the lines OB and I are perpendiculars.

Therefore, by using the definition of perpendicular lines, the measure of the angle$∠BOE$ is 90°.

That implies, $m∠BOE=90°$.

By using the angle addition postulate:

$m∠BOE=m∠4+m∠3$

By using the relation $m∠BOE=90°$, it can be obtained that:

$90°=m∠4+m∠3\left(1\right)$

Substitute t for$m∠3$ into the equation (1).

$$\begin{gathered} 90°=m∠4+m∠3 \\ 90°=m∠4+t \\ 90°−t=m∠4 \end{gathered}$$

Therefore, the measure of the angle ∠4 is $90°−t$.

As $\stackrel{→}{OA}âŠ\yen \stackrel{→}{OC}$,therefore the lines OA and OC are perpendicular.

Therefore, by using the definition of perpendicular lines, the measure of the angle∠AOC is 90° .

That implies, $m∠AOC=90°$.

By using the angle addition postulate:

$m∠AOC=m∠5+m∠4$

By using the relation $m∠AOC=90°$, it can be obtained that:

$90°=m∠5+m∠4\left(2\right)$

Substitute $90°−t$for $m∠4$into the equation (2).

$$\begin{gathered} 90°=m∠5+m∠4 \\ 90°=m∠5+90°−t \\ 90°−90°+t=m∠5 \\ t=m∠5 \end{gathered}$$

Therefore, the measure of angle ∠5 is t.

As $\stackrel{→}{OB}âŠ\yen l$,therefore the lines OB and I are perpendiculars.

Therefore, by using the definition of perpendicular lines, the measure of the angle ∠BOD is 90°.

That implies, $m∠BOD=90°$.

By using the angle addition postulate:

$m∠BOD=m∠5+m∠6$

By using the relation $m∠BOD=90°$, it can be obtained that:

$90°=m∠5+m∠6\left(3\right)$

Substitute t for m∠5into the equation (3).

$$\begin{gathered} 90°=m∠5+m∠6 \\ 90°=t+m∠6 \\ 90°−t=m∠6 \end{gathered}$$

Therefore, the measure of angle ∠6 is $90°−t$.