Question

Help in maths question!

Answer 1

(i) Suppose ∠ADB =∠ x. Since a triangle's angles sum up to 180°, we know that x+α+β=180°. We also know that x+∠ADC=180, so therefore α+β=∠ADC

(ii) The same logic that a triangle's interior angles equal 180°, we know that α+α+β+∠DAC=180. Therefore, 180-α-α-β=∠DAC.

(iii) We know that ∠DAE = 180-2α+β, which means 180-∠DAE = 2α+β, which means (2α+β)/2 which finally means α+β/2, which we will call p. We will call ∠EDC k. Now we know that p+k=α+β, which means k=α+β-p, then α+β-p, but p = (α+β/2), so we plug that in. Finally, everything can be simplified into 3/2*p.

Answer 2

(i) Suppose ∠ADB =∠ x. Since a triangle's angles sum up to 180°, we know that x+α+β=180°. We also know that x+∠ADC=180, so therefore α+β=∠ADC

(ii) The same logic that a triangle's interior angles equal 180°, we know that α+α+β+∠DAC=180. Therefore, 180-α-α-β=∠DAC.

(iii) We know that ∠DAE = 180-2α+β, which means 180-∠DAE = 2α+β, which means (2α+β)/2 which finally means α+β/2, which we will call p. We will call ∠EDC k. Now we know that p+k=α+β, which means k=α+β-p, then α+β-p, but p = (α+β/2), so we plug that in. Finally, everything can be simplified into 3/2*p.

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