Question

6-8a
skills objective d
in 1-10, find the measure of the indicated angles.

1. ( mangle ble = )
2. ( mangle elu = )
3. ( mangle r = )
4. ( mangle red = )
5. ( mangle pnk = )
6. ( mangle inp = )
7. ( mangle pin = )
8. ( mangle bkl = )
9. ( mangle bca = )
10. ( mangle lac = )

(there are some handwritten notes and geometric figures beside the questions, including a right triangle, an isosceles triangle, and other angle - related figures.)

Explanation:

Step1: Solve \( m\angle BLE \)

In triangle \( EBL \), \( \angle B = 90^\circ \), \( \angle E = 56^\circ \). The sum of angles in a triangle is \( 180^\circ \), so \( m\angle BLE=180^\circ - 90^\circ - 56^\circ = 34^\circ \)? Wait, no, wait. Wait, \( \angle BLE \) is inside the right triangle? Wait, no, the first diagram: triangle \( EBL \) is right-angled at \( B \), so \( \angle B = 90^\circ \), \( \angle E = 56^\circ \), so \( \angle BLE = 180 - 90 - 56 = 34^\circ \)? But the original has a wrong answer written. Wait, maybe I misread. Wait, the problem is to find \( m\angle BLE \). Wait, in triangle \( EBL \), right-angled at \( B \), so angles sum to \( 180 \). So \( 90 + 56 + \angle BLE = 180 \), so \( \angle BLE = 34^\circ \). But the written answer is \( 180 \), which is wrong. Let's do it correctly.

Step1: For \( m\angle BLE \)

Triangle \( EBL \) is right - angled at \( B \) (\( \angle B = 90^\circ \)), \( \angle E = 56^\circ \). By the angle - sum property of a triangle (\( \angle B+\angle E+\angle BLE = 180^\circ \)):
\( 90^\circ+56^\circ + m\angle BLE=180^\circ \)
\( m\angle BLE = 180^\circ-(90^\circ + 56^\circ)=34^\circ \)

Step2: For \( m\angle ELU \)

\( \angle BLE \) and \( \angle ELU \) are supplementary (they form a linear pair). So \( m\angle ELU=180^\circ - m\angle BLE \)
Substitute \( m\angle BLE = 34^\circ \), we get \( m\angle ELU = 180^\circ-34^\circ = 146^\circ \)

Step3: For \( m\angle R \) (in triangle \( RDE \))

Triangle \( RDE \) has \( \angle D = 135^\circ \), and since it's an isosceles triangle (from the diagram, \( RD = ED \) maybe? So \( \angle R=\angle E \)). By angle - sum property: \( \angle R+\angle E+\angle D = 180^\circ \), and \( \angle R=\angle E \), so \( 2\angle R+135^\circ = 180^\circ \)
\( 2\angle R=180^\circ - 135^\circ = 45^\circ \)
\( m\angle R=\frac{45^\circ}{2}=22.5^\circ \)

Step4: For \( m\angle RED \)

Since \( \angle R=\angle E = 22.5^\circ \), \( m\angle RED=\angle E = 22.5^\circ \)? Wait, no, \( \angle RED \) is \( \angle E \), which is equal to \( \angle R \) in the isosceles triangle. Wait, the angle - sum property: \( \angle R+\angle E+\angle D = 180^\circ \), and \( \angle R=\angle E \), so \( \angle RED=\angle E = 22.5^\circ \)

Step5: For \( m\angle PNK \)

Assuming \( \angle PNK \) and the \( 128^\circ \) angle (from the diagram, \( \angle PNK \) and the angle marked \( 128^\circ \) are supplementary? Wait, the diagram has an angle of \( 128^\circ \) at \( P \). Wait, maybe \( \angle PNK \) and the adjacent angle are supplementary. Wait, if there is a straight line, and an angle of \( 128^\circ \), then \( m\angle PNK = 180^\circ - 128^\circ=52^\circ \)? Wait, maybe more context. But let's assume the diagram: if there is a linear pair, and one angle is \( 128^\circ \), then \( m\angle PNK = 52^\circ \)? Wait, maybe the diagram shows that \( \angle PNK \) and a \( 128^\circ \) angle are supplementary. So \( m\angle PNK = 180 - 128 = 52^\circ \)

Step6: For \( m\angle INP \)

If \( \angle PNK = 52^\circ \) and \( \angle INK = 112^\circ \) (from the diagram), maybe \( \angle INP \) is supplementary to \( \angle INK \)? Wait, no, maybe \( \angle INP = 180^\circ - 112^\circ = 68^\circ \)? Wait, this is getting a bit unclear without full diagram details, but let's proceed with standard angle - sum and supplementary angle rules.

Step7: For \( m\angle PIN \)

In triangle \( PIN \), if we know two angles, we can find the third. But without full info, but assuming standard problems, if \( \angle INP = 68^\circ \) and \( \angle PNK = 52^\circ \), maybe \( m\angle PIN = 180-(68 + 52)=60^\circ \)? No, this is speculative. Wait, maybe the original problem has standard values.

Step8: For \( m\angle BKL \)

Assuming triangle \( BKL \) or related. If \( \angle B = 42^\circ \) and another angle, but the diagram has some marks. Wait, the last diagram: triangle \( ABC \), with \( \angle B = 42^\circ \), \( \angle BKL \) - maybe \( \angle BKL = 180^\circ - 42^\circ - 32^\circ=106^\circ \)? (assuming some angles)

Step9: For \( m\angle BCA \)

In triangle \( ABC \), if \( \angle B = 42^\circ \) and \( \angle A = 32^\circ \) (from diagram marks), then \( m\angle BCA=180-(42 + 32)=106^\circ \)? No, wait, maybe \( \angle BCA = 180 - 42 - 32 = 106^\circ \)

Step10: For \( m\angle LAC \)

If \( \angle BAC = 32^\circ \) and \( \angle LAC \) is supplementary or related, but this is getting too vague. However, the first two problems:

Answer:

1. \( m\angle BLE=\boldsymbol{34^\circ} \)
2. \( m\angle ELU=\boldsymbol{146^\circ} \)
3. \( m\angle R=\boldsymbol{22.5^\circ} \)
4. \( m\angle RED=\boldsymbol{22.5^\circ} \)

(For the rest, more diagram - specific info is needed, but the first four are solved with standard triangle angle - sum and supplementary angle rules)