2026 April(R4A) Math Question 26
Trigonometry
- Pythagorean Theorem & sin cos tan
With the right angle at BBB, BC‾\overline{BC}BC is opposite ∠A\angle A∠A and AC‾\overline{AC}AC is adjacent to ∠A\angle A∠A. tan70∘=BCAC=9AC\tan 70^\circ = \frac{BC}{AC} = \frac{9}{AC}tan70∘=ACBC=AC9 AC=9tan70∘AC = \frac{9}{\tan 70^\circ}AC=tan70∘9
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