via)

Triangle $A O B$ is a right-angled triangle, use SOHCAHTOA to find length $O B$

Label the sides of the triangle: hypotenuse, adjacent and opposite

Circle theorem question (labelled for SOHCAHTOA)

You know the adjacent side and want to calculate the opposite side, so use the tangent ratio, $\tan θ = \frac{O}{A}$

Substitute in the values that you know

$\tan 49 = \frac{O B}{5.4}$

[1]

Multiply both sides by 5.4 to find length $O B$

$5.4 \times \tan 49 = O B$

$O B = 6.21198 . . .$

Round

$O B = 6.21 cm$ [1]

Answers between 6.211 and 6.212 will be accepted

vib)

Use Pythagoras' theorem, $a^{2} + b^{2} = c^{2}$ to calculate the length $O A$
It may help you to label the triangle first

Circle theorem question (labelled fro Pythagoras' theorem)

Substitute in the lengths that you know, use the unrounded value for $O B$

$5 . 4^{2} + 6.21198 . . .^{2} = O A^{2}$

[1]

Take the square root of both sides

$\sqrt{5 . 4^{2} + 6.21198 . . .^{2}} = O A$

Evaluate

$O A = 8.23096 . . .$

Round

$O A = 8.23 cm$ [1]

Answers between 8.229 and 8.231 will be accepted

vic)

Use the formula for the area of a triangle, $A = \frac{1}{2} \times base \times height$
Use the unrounded value of $O B$

$A = \frac{1}{2} \times 5.4 \times 6.21198 . . .$

[1]

$A = 16.77237 . . .$

Round

$16.8 {cm}^{2}$ [1]

Answers between 16.76 and 16.77 will be accepted

Circle Theorems (CIE IGCSE Maths: Core)

Exam Questions