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° m∠cdb = m∠cbd = m∠bcd = m∠cda = m∠ cad= m∠acd =

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°

step-by-step explanation: a reaction is first order. if the initial reactant concentration is 0.0200 m, and 25.0 days later the concentration is 6.25 x 10-4 m, then its half-life is: