MATH6058 Solve for x and y on a right angled triangle

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Question 1

Solve the following mensuration questions to 2 decimal places:

Note: You may need to scroll down to see the diagrams.

a)

Find the values of x and y:

Note:The bottom left angle is a right angle (90 $^{\circ}$ ).

x = Round your answer to 2 decimal places.Expected answer:mm

y = $^{\circ}$

To find x: Use Pythagoras Theorem to find the missing side.

To find y: For the most accurate answer use the sides that were given. What trigonometric ratio includes those two sides? Sin. Remember to use inverse-sin to find the angle.

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b)

Find the values of x and y:

Note:The top right angle is a right angle (90 $^{\circ}$ ).

x = $^{\circ}$

y = Round your answer to 2 decimal places.Expected answer: mm

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Advice

i) $\frac{\var{lent21}}{\sin(90^{\circ})} = \frac{\var{lent22}}{\sin(Y^{\circ})}$

$\sin(Y^{\circ}) = \var{lent22} \div \frac{\var{lent21}}{\sin(90^{\circ})}$

$Y = \var{ans21}^{\circ}$

$\frac{\var{lent21}}{\sin(90^{\circ})} = \frac{x}{\sin(90^{\circ}-\var{ans21})}$

$x = \frac{\var{lent21}}{\sin(90^{\circ})} \times \sin(90^{\circ}-\var{ans21}^{\circ}) = \var{ans22}mm$

ii) $y^2 + \var{lent11}^2 = \var{lent12}^2$

$y^2 = \var{lent12}^2 - \var{lent11}^2$

$y = \sqrt(\var{lent12}^2 - \var{lent11}^2) = \var{ans11}mm$

$\frac{\var{lent12}}{\sin(90^{\circ})} = \frac{\var{ans21}}{\sin(x^{\circ})}$

$\sin(x^{\circ}) = \var{ans21} \div \frac{\var{lent12}}{\sin(90^{\circ})}$

$x = \var{ans12}^{\circ}$

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MATH6058 Solve for x and y on a right angled triangle

a)

b)

Advice