<h2>
54 units²</h2><h2 />
This is a compound shape. You can split it into x shapes. See Attachment
Area of a Rectangle = L × B
L = 7
B = 6
7 × 6 = 42
<h3>42 units²</h3>
Area of a Triangle = 1/2BH
B = 6
H = 1
1/2 × 6 × 1 = 3
<h3>3 units²</h3>
Area of a Triangle = 1/2BH
B = 3
H = 2
1/2 × 3 × 2 = 3
<h3>3 units²</h3>
Area of a Triangle = 1/2BH
B = 2
H = 6
1/2 × 2 × 6 = 6
<h3>6 units²</h3><h3 /><h3>42 + 6 + 3 + 3 = 54</h3>
A rhombus diagonals form right angles. The diagonals also cut in half the angles
use cosine: cos38 = 
a = 20cos38
a=15.8
Answer:
<u>Identities used:</u>
- <em>1/cosθ = secθ</em>
- <em>1/sinθ = cosecθ</em>
- <em>sinθ/cosθ = tanθ</em>
- <em>cosθ/sinθ = cotθ</em>
- <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
- (1 - sinθ)/(1 + sinθ) =
- (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
- (1 - sinθ)² / (1 - sin²θ) =
- (1 - sinθ)² / cos²θ
<u>Square root of it is:</u>
- (1 - sinθ)/ cosθ =
- 1/cosθ - sinθ / cosθ =
- secθ - tanθ
<h3>Question 2 </h3>
<u>The first part without root:</u>
- (1 + cosθ) / (1 - cosθ) =
- (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
- (1 + cosθ)² / (1 - cos²θ) =
- (1 + cosθ)² / sin²θ
<u>Its square root is:</u>
- (1 + cosθ) / sinθ =
- 1/sinθ + cosθ/sinθ =
- cosecθ + cotθ
<u>The second part without root:</u>
- (1 - cosθ) / (1 + cosθ) =
- (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
- (1 - cosθ)²/ (1 - cos²θ) =
- (1 - cosθ)²/sin²θ
<u>Its square root is:</u>
- (1 - cosθ) / sinθ =
- 1/sinθ - cosθ / sinθ =
- cosecθ - cotθ
<u>Sum of the results:</u>
- cosecθ + cotθ + cosecθ - cotθ =
- 2cosecθ
Answer:
A. Undefined.
Step-by-step explanation:
There is no real square root of a negative number.
