(a) Prove that the sum of the angles in a triangle... | WAEC SSCE - General Mathematics - 1988 (Essay)

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(a) Prove that the sum of the angles in a triangle is two right angles.

(b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ML produced at Q. If < LMN = 46°, and < MLN = 80°, calculate < LQN, stating clearly your reasins.

Answer

(a) The angle sum of a triangle is two right angles.

Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).

\(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
\(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).

Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)

(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).

\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).

Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).

In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)

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Answer Details

(a) The angle sum of a triangle is two right angles.

Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).

\(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
\(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).

Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)

(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).

\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).

Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).

In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)

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(a) Prove that the sum of the angles in a triangle is two right angles. (b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ...

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(a) Prove that the sum of the angles in a triangle is two right angles. (b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ...

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