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2 years ago

15

PLS HELP / EXPLAIN !!!!

1 answer:

Scorpion4ik [409]2 years ago

5 0

Area of a triangle = $\frac{1}{2}$ · base x height
   180 = $\frac{1}{2}$ · base x 40
   180 = 20 · base
      9 = base

Pythagorean Theorem:
base² + height² = hypotenuse²
9²    + 40² = hypotenuse²
81 + 1600 = hypotenuse²
1681    = hypotenuse²
   √1681           = √hypotenuse²
   41    = hypotenuse

Answer: the length of the hypotenuse is 41 m

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Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ .

The sine difference identity is:

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a)$ .

The cosine addition identity is:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ .

The cosine difference identity is:

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ .

We need to find a way to put some or all of these together to get:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$ .

So I do notice on the right hand side the $\sin(a+b)$ and the $\sin(a-b)$ .

Let's start there then.

There is a plus sign in between them so let's add those together:

$\sin(a+b)+\sin(a-b)$

$=[\sin(a+b)]+[\sin(a-b)]$

$=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]$

There are two pairs of like terms. I will gather them together so you can see it more clearly:

$=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]$

$=2\sin(a)\cos(b)+0$

$=2\sin(a)\cos(b)$

So this implies:

$\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$

Divide both sides by 2:

$\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)$

By the symmetric property we can write:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$

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Answer:

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Step-by-step explanation:

-6p = 48

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