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Question

A triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that $$\angle BAL=\angle ACB$$.


Solution

In $$\triangle ABL$$,
$$\angle BAL+\angle ALB+\angle B=180^{o}$$
$$\angle BAL+90^{o}+\angle B=180^{o}$$
$$\angle BAL=90^{o}-\angle B$$          ......(1)
In $$\triangle ABC$$,
$$\angle A +\angle B +\angle C=180^{o}$$
$$\angle B+\angle C=90^{o}$$
$$\angle C=90^{o}-\angle B$$
$$\angle ACB=90^{o}-\angle B$$          .......(2)
From 1 and 2, we get,
$$\angle BAL=\angle ACB$$

1333545_1111013_ans_e3898781482d40f381231093a84b3b55.png

Mathematics

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