Steps for construction:
Step 1: Draw the rough diagram and mark the measurements.
Step 2: Draw the ΔABC with the given measurements.
Step 3: Construct altitudes from any two vertices A and C to their opposite sides BC and AB respectively.
Step 4: The point of intersection of the altitude H is the orthocentre of the given ΔABC.

Steps for construction:
Step 1: Draw the ΔPQR with the given measurements.
Step 2: Construct altitudes from any two vertices P and Q to their opposite sides QR and PR respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given  ΔPQR.

Step 1: Draw the ΔPQR with the given measurements


Step 2:
Construct altitudes from any two vertices (say) R and P, to their opposite sides PQ and QR respectively.
The point of intersection of the altitude H is the Orthocentre of the given ΔPQR.

Given PQ = PR
∴ ㄥR = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (QR arid PR) and let them meet at S. S is the circumcenter of Δ PQR.
Step 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.


Circum radius = 3.5 cm.
Construction of Orthocentre of a Triangle:
Orthocentre: The orthocentre is the point of concurrency of the altitudes of a triangle. Usually it is denoted by H.

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, Band C.


Circum radius = 4.3 cm.

Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.


Circum radius = 4.3 cm.

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.


Circum radius = 4.3 cm.

\(\angle A+\angle B=\angle ACD\) (An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x - 5 = 4x - 15
3x + 30 = 4x- 15
30+15 = 4x-3x
45° = x
\(\angle A=x+35^0\)
= 45°+ 35°
= 80°
\(\angle B=2x-5\)
= 2(45°) - 5°
= 90° - 5° = 85°
\(\angle ACD=4x-15\)
= 4(45°)-15°
= 180°-15° = 165°
\(\angle ACB=180^0-\angle ACD\)
=180° - 165° = 15°
\(\angle A=80^0,\angle B=85^0,\angle C=15^0\).

In \(\triangle ABC\angle B=60^0\ and\ \angle C=70^0\)
\(\therefore\angle A=180^0-(60^0+70^0)\)
\(=180^0-130^0\) = 50⁰
In \(\triangle E=70^0\ and \angle D=60^0\)
\(\therefore \angle F=180^0-(70^0+60^0)\)
= 180° - 130° = 50°
\(\angle A=\angle F=50^0\)
\(\angle B=\angle D=60^0\)
\(\angle C=\angle E=70^0\)
By AAA congruency
\(\therefore\triangle ABC\cong\triangle FDE\)

The ratio of the angles of a triangle = 1: 2 : 3.
\(\therefore\) Let the angles of a triangle be x, 2x and 3x.
x + 2x + 3x = 180° (Total angle of a triangle is 180°)
6x = 180°
\(x=\frac{180^0}{6}=30^0\)
x = 30° ; 2x = 2x30° = 60° ; 3x = 3 x 30° = 90°
Measures of the angles of a triangle = 30°, 60° and 90°.