IGCSE Trigonometry 0580 Exam 2025 Preparation

Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended

Practice problems (IGCSE Trigonometry 0580 Exam 2025 preparation)

Learn to find the missing Length of the Side of a right Triangle in Trigonometry

Finding the unknown Length of the Side of a right Triangle in Trigonometry

Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended)

Q1 Work out the length of CD. Give your answer correct to 3 sf.

Solution: We have to find side CD. But in right triangle ∆BDC, only one angle 40° is known. We require one given side also.

So, from ∆ABD, we will try to find side BD. Now ∆ABD is right triangle, so we can apply Pythagoras theorem to find the side BD.

Using Pythagoras Theorem

H²=P²+B²

16²=BD²+12²

BD²=16²-12²=256-144=112

BD=√112=10.58

In right triangle ∆BDC,

Sin 40°=O/H=BD/CD=10.58/CD

CD=10.58/sin 40°=10.58/0.64=16.53

= 16.5 cm

Q2 Work out the length of AC. Give your answer correct to 3 sf.

Solution: We have to find side AC. Here AC is sum of two sides AD and DC. So, we have to find these two sides from ∆ABD and ∆BDC respectively.

In right triangle ABD,

Sin 65°=O/H=BD/AB=BD/7

BD=7× sin 65°=6.34 cm

Cos 65°=A/H=AD/AB=AD/7

AD = 7× cos 65° =2.96

From ∆BDC, we can apply Pythagoras theorem to find the side CD.

CD²=BC²-BD²=12²-6.34²=103.8

CD=10.18

AC= AD + DC= 2.96 + 10.18=13.14 or 13.1 (3 sf)

Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended)

Q3 Calculate the length x. Give your answer correct to 3 sf.

Solution: We have to find side BD. But in right triangle ∆BDC, only one angle 25° is known. We require one given side also.

So, from ∆ABC, we will try to find side BC. Now ∆ABC is right triangle, so we can apply Pythagoras theorem to find the side BC.

Using Pythagoras Theorem

H²=P²+B²

20²=BC²+16²

BC²=20²-16²=400-256=144

BC=√144=12

In right triangle ∆BDC,

Sin 25°=O/H=BC/BD=12/BD

BD=12/sin 25°=28.39=28.4 (3sf)

Q4 Given BD=1 m, calculate the length AC.

Solution: We have ∆ABD and ∆ACD, two right triangles. Now we have to find side AC, and this side is part of ∆ACD. But in this triangle only one angle 21° is given to us. We need one more side. We can observe that from right ∆ABD we can find side AD.

So, using ∆ABD, let’s find side AD.

From ∆ABD, Cos 21°=A/H=AD/BD=AD/1

AD=cos 21°=0.93

Now in ∆ACD, sin 21°= O/H=AC/AD=AC/0.93

AC=0.93× sin 21°=0.3332

AC=0.333 m

Q5 In the triangle PQR, ∟ PQR=90° and ∟RPQ=31°. The length of PQ is 11 cm. Calculate

the length of the perpendicular from Q to PR

Solution: Let QM is perpendicular to PR.

In ∆PMQ, sin 31°=O/H=QM/PQ=QM/11

QM=11×sin 31° = 5.67 cm

Q6 Calculate the length of AB

Solution: We have to find side AB. But in right triangle ∆ABD, only one angle 35° is known. We require one given side also.

So, from ∆BCD, we will try to find side BD.

Now ∆BCD is right triangle

Cos 20°=A/H=BD/10

BD=10×cos 20°=9.40

In right triangle ∆ABD,

Sin 35°=O/H=AB/BD=AB/9.4

AB=9.4×sin35°=5.39 cm

Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended)