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°
Answer:
10.5 : 12 : 13.5
Step-by-step explanation:
7:8:9
We need to get 8 to 12
12/8 = 3/2
We need to multiply each number by 3/2
7 * 3/2:8* 3/2 :9*3/2
21/2 : 12 : 27/2
10.5 : 12 : 13.5
Answer: 7(3m - 7n)
<u>Step-by-step explanation:</u>
Factor each term then see what value(s) they have in common:
21m - 49n
∧ ∧
3 · 7 · m 7 · 7 · n
Both terms have a 7 in common. What is left over after removing the 7?
21m: 7(3m)
49n: 7(7n)
GCF: 7 Leftovers: 3m <em>and</em> 7n
Factored form of expression is: 7(3m - 7n)