A five-card poker hand is dealt at random from a standard 52-card deck. Note the total number of possible hands is C(52,5)=2,598
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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°
First multiply each term by 24(the common denominator found by 8×3) to remove the fractions and make things easier.
This will give you,
Then just continue to simplify and isolate the variable.
This will give you,
Then just continue to simplify and isolate the variable.