In ΔABC, AB = x units AC = 7 units BC = √(49 - x²) =√[(7+x)(7-x)] units by Pythagoras thm. ∠A+∠C = 90° => Cos A = Sin C and Cot A = Tan C
Tan C = Cot A = AB/BC = x/√[(7+x)(7-x)] Cos A = Sin C = x/7
LHS = [ √(7-x) + √(7+x) ] * Tan C + Cos A (-14 +21) + √(49+x²) * Cos 90° = [√(7-x) + √(7+x)] * x/√[49-x²) + 7 * x/7 + √(49+x²) * 0 = x [ √(7-x) + √(7+x)] / √(49-x²) + x If required, Rationalize the denominator in the first term by multiplying Nr and Dr by √(49-x²).
looks like we cant simplify this further. ===============
ΔPQR: PQ : QR : PR = 8 : 15 : 17 check that PR² = PQ² + QR² So it is a right angle triangle , they make a Pythagorean triplet. As PR is largest, it is Hypotenuse. ∠Q = 90°. ∠P + ∠R = 90° Draw the triangle with this description.
Sin R = Cos P = PQ/PR = 8/17 Sin P = Cos R = QR/PR = 15/17
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In ΔABC, AB = x units AC = 7 unitsBC = √(49 - x²) =√[(7+x)(7-x)] units by Pythagoras thm.
∠A+∠C = 90° => Cos A = Sin C and Cot A = Tan C
Tan C = Cot A = AB/BC = x/√[(7+x)(7-x)]
Cos A = Sin C = x/7
LHS = [ √(7-x) + √(7+x) ] * Tan C + Cos A (-14 +21) + √(49+x²) * Cos 90°
= [√(7-x) + √(7+x)] * x/√[49-x²) + 7 * x/7 + √(49+x²) * 0
= x [ √(7-x) + √(7+x)] / √(49-x²) + x
If required, Rationalize the denominator in the first term by multiplying Nr and Dr by √(49-x²).
looks like we cant simplify this further.
===============
ΔPQR:
PQ : QR : PR = 8 : 15 : 17
check that PR² = PQ² + QR²
So it is a right angle triangle , they make a Pythagorean triplet. As PR is largest, it is Hypotenuse. ∠Q = 90°. ∠P + ∠R = 90°
Draw the triangle with this description.
Sin R = Cos P = PQ/PR = 8/17
Sin P = Cos R = QR/PR = 15/17
LHS = 0 as Cos P cosR = sin P sin R.
Step-by-step explanation:
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