Question

In the right triangle shown, m\angle Q = 60\degreem∠Q=60°m, angle, Q, equals, 60, degree and QR=2\sqrt 3QR=2
3
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Q, R, equals, 2, square root of, 3, end square root.

Answer 1

Answer:

$4\sqrt{3$ is the answer then looking at this question.

Answer 2

Concept:

We need to know how to solve sin∅ and cos∅

$sin\theta =\frac{opposite}{hypotenuse}\\cos\theta =\frac{adjacent}{hypotenuse}$

Also, sin 60°=$\frac{\sqrt{3} }{2}$ and cos 60°=$\frac{1}{2}$.

Given:

It is given that ΔPRQ is right angled at R, value of ∠Q is 60° and length QR is $\frac{2}{\sqrt{3} }$

To find:

We need to find the length of PQ (hypotenuse)

Solution:

Consider, in ΔPQR length of PQ=x

Now cos∠Q=cos60°

$cos60=\frac{{1} }{2}$                 ------------(1)

Also, $cos\angle Q=\frac{QR}{x}$                   -----------(2)

From equation (1) and (2)

$\frac{1}{2}=\frac{QR}{x}$

Substituting value of QR

$\frac{1}{2}= \frac{\frac{2}{\sqrt{3} } }{x}$

⇒$\frac{1}{2} =\frac{2}{\sqrt{3} }\times \frac{1}{x}$

⇒$x=\frac{2\times 2}{\sqrt{3} }=\frac{4}{\sqrt{3} }$

⇒$x=\frac{4\sqrt{3} }{\sqrt{3} \times \sqrt{3} }=\frac{4\sqrt{3} }{3}$

Therefore, the value of PQ is $\frac{4\sqrt{3} }{3}$cm.