<em>The answer is</em><em> $300</em><em>......</em>
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
Answer:
32°
Step-by-step explanation:
Given:
∠DMQ = 58º
In this circle, the radius is DM. Since AD is tangent to the circle M, at point D, and the angle between a tangent and a radius is 90°
Therefore, ∠MDQ = 90°
The total angle in a triangle is 180°. Since we have the values of ∠MDQ and ∠DMQ, ∠DQM will be calculated as:
180 = ∠DMQ + ∠MDQ + ∠DQM
Solving for ∠DQM, we have:
∠DQM = 180 - ∠DMQ - ∠MDQ
∠DQM = 180 - 90 - 58
∠DQM = 32°
The measure of ∠DQM is 32°
The answer is A trust me I did it before