In trigonometry, the right triangle is considered a special triangle because there are derived equations solely for this type. It is really convenient when dealing right triangle problems because it is more simplified courtesy of the Pythagorean theorems. It is derived that the square of the hypotenuse (longest side of the triangle) is equal to the sum of the squares of the other two legs. In equation, that would be c² = a² + b². For this activity, all you have to do is find the sum of the squares in columns a and b. Then, see if this is equal to the square of the values in column c. Let's calculate each row:
Row 1:
3² + 4² ? 5²
25 ? 25
25 = 25
Row 2:
5² + 12² ? 13²
169 ? 169
169 = 169
Row 3:
9² + 12² ? 15²
225 ? 225
225 = 225
Therefore, all of the given values conform to a² + b² = c².
Numerator
<span><span>cos<span>(<span>π/2</span>−x)</span></span>=<span>cos<span>(<span>π/2</span>)</span></span><span>cosx</span>+<span>sin<span>(<span>π/2</span>)</span></span><span>sinx</span></span>
now <span><span>cos<span>(<span>π/2</span>)</span></span>=0 and <span>sin<span>(<span>π/2</span>)</span></span>=1</span>
simplifies to : 0 + sinx = sinx
Denominator
<span><span>sin<span>(<span>π/2</span>−x)</span></span>=<span>sin<span>(<span>π/2</span>)</span></span><span>cosx</span>+<span>cos<span>(<span>π/2</span>)</span></span><span>sinx</span></span>
simplifies to : cosx + 0 = cosx
<span>⇒<span><span>cos<span>(<span>π/2</span>−x)</span></span><span>sin<span>(<span>π/2</span>−x)</span></span></span>=<span><span>sinx/</span><span>cosx</span></span>=<span>tan<span>x</span></span></span>
It is C: Given, Reflexive Property
Answer:
12s
Step-by-step explanation:
5s + 7s = (5 + 7) s= 12s
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