Question
Construct with ruler and compasses, angles of measure: $60^\circ $
✓
Answer
The below given steps will be followed to construct an angle of $60^\circ .$
$1.$Draw a line $l$ and mark a point $P$ on it. Now, taking $P$ as centre and with a convenient radius, draw an arc of a circle which intersects line $l$ at $Q.$
$2.$Taking $Q$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $R.$
$3.$Join $PR$ which is the required ray making $60^\circ $ with line $l.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Construct a right angle $A B C$ with the help of a protractor. Mark a point $D$ in the interior of $\angle A B C$. Draw ray $B D$. Measure $\angle A B D, \angle D B C$. Verify that they are complementary.
→Construct an angle of $120^\circ $ and bisect it.
→Complete the following table:
From the above table:
(i) Write all the pairs of integers whose sum is 0.
(ii) Is (-4) + (-2) = (-2) + (-4)?
(iii) Is 0 + (-6) = -6?
| + | -6 | -4 | -2 | 0 | 2 | 4 | 6 |
| 6 | 10 | ||||||
| 4 | |||||||
| 2 | 8 | ||||||
| 0 | -6 | ||||||
| -2 | |||||||
| -4 | 0 | ||||||
| -6 | -6 |
(i) Write all the pairs of integers whose sum is 0.
(ii) Is (-4) + (-2) = (-2) + (-4)?
(iii) Is 0 + (-6) = -6?
Draw a right angle and construct its bisector.
Draw an angle of measure $135^\circ $ and bisect it.
→→
Use unmarked number lines to evaluate these expressions:
(a) -125 + (-30) = _________
(b) +105 – (-55) = _________
(c) +80 – (-150) = _________
(d) -99 – (-200) = _________
(a) -125 + (-30) = _________
(b) +105 – (-55) = _________
(c) +80 – (-150) = _________
(d) -99 – (-200) = _________
Find the equivalent fraction of $\frac{2}{9}$ with denominator $63.$
→Draw a line segment $A B$ and bisect it. Bisect one of the equal parts to obtain a line segment of length $\frac{1}{2}(A B)$.
→Draw any line segment $\overline{\text{PQ}}$. Without measuring $\overline{\text{PQ}}$, construct a copy of $\overline{\text{PQ}}$.
→