What is the height of a triangle when the distance between angles 62 degrees and 48 degrees is 1.8 cm?

What is the height of a triangle when the distance between angles 62 degrees and 48 degrees is 1.8 cm?
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There is probably a trick that I don't know (can't think of atthe moment), but you can use the sine rule and sine ratio:The third angle is 180° - (62° + 48°) = 70° and is opposite theside of length 1.8cm.The side opposite the 48° can be found using the sine rule:a/sin A = b/sin B → a = b × sin A/sin B = 1.8 cm × sin48° / sin70°The height can now be found using the sine ratio on the 62°angle as the side just found is the hypotenuse of thattriangle:sine = opp/hyp→ opp = hyp × sine→ height = (1.8 cm × sin48° / sin 70°) × sin 62°→ height = 1.8 × sin 48° × sin 62° / sin 70° cm ≈ 1.3 cm
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