8.2.2 Lines and Angles, PT3 Practice - Mathematics

Question 4:

In Diagram below, PQR is a straight line. Find the value of x.

Solution:

\begin{array}{l} x^{o} = ∠ Q R T + ∠ T R S \\ ∠ Q R T = 40^{o} \\ ∠ T R S = 180^{o} - 135^{o} \\ = 45^{o} \\ Hence, x^{o} = 40^{o} + 45^{o} \\ x^{o} = 85^{o} \\ x = 85 \end{array}

$\begin{array}{l} x^{o} = ∠ Q R T + ∠ T R S \\ ∠ Q R T = 40^{o} \\ ∠ T R S = 180^{o} - 135^{o} \\ = 45^{o} \\ Hence, x^{o} = 40^{o} + 45^{o} \\ x^{o} = 85^{o} \\ x = 85 \end{array}$

Question 5:
In Diagram below, find the value of y.

Solution:

\begin{array}{l} ∠ A B C = ∠ B C D \\ = 108^{o} (alternate angle) \\ y^{o} + 130^{o} + 108^{o} = 360^{o} \\ y^{o} = 360^{o} - 238^{o} \\ y^{o} = 122^{o} \end{array}

$\begin{array}{l} ∠ A B C = ∠ B C D \\ = 108^{o} (alternate angle) \\ y^{o} + 130^{o} + 108^{o} = 360^{o} \\ y^{o} = 360^{o} - 238^{o} \\ y^{o} = 122^{o} \end{array}$

Question 6:
In Diagram below, PSR and QST are straight lines.

Find the value of x.

Solution:

\begin{array}{l} ∠ U S T + ∠ S T V = 180^{o} \\ ∠ U S T = 180^{o} - 116^{o} = 64^{o} \\ ∠ P S T = ∠ Q S R \\ x^{o} + ∠ U S T = 135^{o} \\ x^{o} + ∠ U S T = 135^{o} \\ x^{o} = 71^{o} \\ x = 71 \end{array}

$\begin{array}{l} ∠ U S T + ∠ S T V = 180^{o} \\ ∠ U S T = 180^{o} - 116^{o} = 64^{o} \\ ∠ P S T = ∠ Q S R \\ x^{o} + ∠ U S T = 135^{o} \\ x^{o} + ∠ U S T = 135^{o} \\ x^{o} = 71^{o} \\ x = 71 \end{array}$

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