IGCSE Trigonometry 0580 Exam 2025 preparation 

Finding an Unknown Angle of a Right Triangle Using Trigonometry

Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended)

Q1 Work out the size of angle x. Given that BC is parallel to AD.

Solution:

To solve this problem, we have to draw CN perpendicular to AD as shown in figure.

Now BCNM becomes a rectangle. Thus BM=CN=6 cm and BC=MN=10 cm.

Now to find angle x, we require side ND also.

From ∆ABM, using Pythagourus Theorem,

AM²=AB²-BM²=7.5²-6²=20.25

AM=√20.25=4.5

Now ND=AD-(AM+MN)=24-(4.5+10)=9.5

Now in ∆CND,

Tan x=O/A=6/9.5=0.632

x=Tan⁻¹ 0.632 =32.3°

Explore the Trigonometry course with free video lectures

 Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended

Q2 Find the angle marked with x.

Solution: In order to find angle x from ∆BCD, we require one more side. We can find side BD from right ∆ABD.

Consider ∆ABD, sin 27.2°=opposite /hypotenuse=BD/100

BD=100 × Sin 27.2°=45.71

In ∆BCD,

Sin x= BD/BC=45.71/85=0.538

X=sin⁻¹ 0.538=32.5°

Explore the Trigonometry course with free video lectures

 Trigonometry Course for Cambridge IGCSE Mathematics (0580) Extended

Q3 In the triangle ∆XYZ, XY = 14 cm, XZ = 17 cm and ∟ YXZ = 25°. A is the foot of the perpendicular from Y to XZ.

 Calculate:

a) the length XA

b) the length YA

c) the ∟ZYA=θ

Solution:

For finding sides XA, YA, consider ∆AXY

Cos 25°=Adjacent/Hypotenuse=XA/14

XA=14 × cos25° =12.69 or 12.7 cm

sin 25°=opposite/Hypotenuse=YA/14

YA=14 × sin 25° =5.91 or 5.9 cm

AZ=XZ-XA=17cm – 12.7 cm=4.3 cm

Since we require ∟θ,

Consider ∆AZY,

AZ/YA=opposite /Adjacent =tanθ

Tan θ = 4.3/5.9=0.73

θ=tan⁻¹(0.73)=36.1°

Answer 12.7, 5.9, 36.1°