Answer:
m∠H = 10°
Step-by-step explanation:
ΔFGH is an isosceles triangle. I made a diagram, so m∠F ≅ m∠H by B.A.T(base angle theorem). This is because an isosceles triangle have to congruent sides and angles.
Diagram is attached:
-Chetan K
Answer:
√(2 + √3)/4
Step-by-step explanation:
Sine 5π/12 = Sine (5π/6)/2
Recall
π = 180°
Thus,
Sine (5π/6)/2 = Sine (5×180 /6)/2
= Sine 150/2
Recall
Sine θ/2 = √(1 – Cos θ)/2
Thus,
Sine 150/2 = √(1 – Cos 150)/2
But, Cosine is negative in the 2nd quadrant. Thus,
Cos 150 = – Cos 30 = –√3/2
Thus,
√(1 – Cos 150)/2 = √(1 – –√3/2 )/2
= √(1 + √3/2 )/2
= √[(2 + √3)/2 ÷ 2]
= √[(2 + √3)/2 × 1/2]
= √(2 + √3)/4
Therefore,
Sine 5π/12 = √(2 + √3)/4
Tan ( A - B ) = ( tan A - tan B ) / ( 1 + tan A tan B )
tan A = 3 tan B/2
tan ( A - B ) = ((3 tan B/ 2)-tan B) / ( 1 + 3 tan² B/2)=
= (tan B/2) / ( 2 + 3 sin²B/cos²B )=
= (sin B / cos B) / (( 2cos² B+3sin²B)/cos²B)=
=( sin B cos B ) / ( 2 cos²B + 3 ( 1 - cos² B ) ) =
= (sin B cos B ) / ( 2 cos² B + 3 - 3 cos² B ) =
= ( sin 2 B ) / 2 ( 3 - cos² B ) =
= ( sin 2 B ) / ( 6 - cos² B )=
= ( sin 2 B ) / ( 5 + 1 - 2 cos² B )=
= ( sin 2 B ) / ( 5 + sin² B + cos ² B - 2 cos² B ) =
= ( sin 2 B ) / ( 5 - ( cos² B - sin² B ) ) =
= ( sin 2 B ) / ( 5 - cos 2 B ) - correct
