In ∆PQR, angle Q =90°, PR= √5, PQ-RQ=1. Find (cosP- cosR)
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In ∆PQR, angle Q =90°, PR= √5, PQ-RQ=1. Find (cosP- cosR)
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We can use the Pythagorean theorem to find QR. Since angle Q is 90°, we have a right triangle. Let's denote QR as x.
According to the Pythagorean theorem, PR^2 = PQ^2 + QR^2.
Substituting the given values, we have:
√5^2 = (PQ - x)^2 + x^2.
Simplifying the equation:
5 = PQ^2 - 2PQx + x^2 + x^2.
Since PQ - RQ = 1, we can substitute PQ - x with 1:
5 = 1^2 - 2x + x^2 + x^2.
Simplifying further:
5 = 1 - 2x + 2x^2.
Now, we can solve this quadratic equation to find the value of x. Once we have x, we can find the angles P and R using trigonometric ratios. Finally, we can calculate cosP and cosR and subtract them to find (cosP - cosR).
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