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umka21 [38]
3 years ago
10

In triangle ΔABC, ∠C is a right angle and CD is the height to AB . Find the angles in ΔCBD and ΔCAD if: Chapter Reference a m∠A

= 20° Answer: m∠CDB = m∠CBD = m∠BCD = m∠CDA = m∠ CAD= m∠ACD =

Mathematics
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

Given A triangle ABC in which

 ∠C =90°,∠A=20° and CD ⊥ AB.

In Δ ABC

⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]

⇒20° + ∠B + 90°=180°

⇒∠B+110° =180°

∠B =180° -110°

∠B = 70°

In Δ B DC

∠BDC =90°,∠B =70°,∠BC D=?

∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]

90° + 70°+∠BC D =180°

∠BC D=180°- 160°

∠BC D = 20°

In Δ AC D

∠A=20°, ∠ADC=90°,∠AC D=?

∠A +  ∠ADC +∠AC D=180° [angle sum property of triangle]

20°+90°+∠AC D=180°

110° +∠AC D=180°

∠AC D=180°-110°

∠AC D=70°

So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°





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