How do you find the base of a right triangle when given the hypotenuse is 14 ft. the angle formed between the hypotenuse and base is 41 degrees?

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Answer 1 · 2016-07-14T03:22:07

$a = 10.57\quad ft$

There's a relationship that exists between the sides of a right triangle. These are called the trigonometric identities. The three basics are:

$sin(\theta)=\frac{o}{h}$

$cos(\theta)=\frac{a}{h}$

$tan(\theta)=\frac{o}{a}=\frac{sin(\theta)}{cos(\theta)}$

There's a useful mnemonic you can use to remember them:

S ine
O pposite
H ypothenuse

C osine
A djacent
H ypothenuse

T angent
O pposite
A djacent

So, given the hypothenuse and the angle formed between it and its base (adjacent side), the relationship formed is the cosine:

$cos(\theta)=\frac{a}{h}$

Since we know the angle $\theta=41^\circ$ and the hypothenuse $h=14ft$ , all we need to do is to solve for its adjacent side $a$ .

$cos(\theta)=\frac{a}{h}$

$a \quad= h*cos(\theta)$
$\qquad= 14 * cos(41)$
$\qquad = 10.57\quad ft$