How do you find two missing sides of a right triangle with a right angle and a 50 degree angle, the hypotenuse is 2√2?

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How do you find two missing sides of a right triangle with a right angle and a 50 degree angle, the hypotenuse is 2√2?

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Answer 1 · 2016-12-31T12:02:34

Missing sides are $2.17 (2 d p), 1.82 (2 d p)$ $2.17(2dp), 1.82(2dp)$ units

Explanation:

Let hypotenuse of right triangle is $c = 2 \sqrt{2}$ $c=2sqrt2$ . Base of right triangle is $a$ $a$ and perpendicular of right triangle is $b$ $b$ and angle between sides $a and c$ $a and c$ is $∠ B = 50^{0}; sin B = \frac{b}{c} or sin 50 = \frac{b}{2 \sqrt{2}} or b = 2 \sqrt{2} sin 50 = 2.17 (2 d p)$ $/_B=50^0 ; sinB=b/c or sin50=b/(2sqrt2) or b=2sqrt2sin50=2.17(2dp)$

$cos B = \frac{a}{c} or cos 50 = \frac{a}{2 \sqrt{2}} or a = 2 \sqrt{2} cos 50 = 1.82 (2 d p)$ $cosB=a/c or cos50=a/(2sqrt2) or a=2sqrt2 cos50 =1.82(2dp)$

Missing sides are $2.17 (2 d p), 1.82 (2 d p)$ $2.17(2dp), 1.82(2dp)$ units [Ans]

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How do you find two missing sides of a right triangle with a right angle and a 50 degree angle, the hypotenuse is 2√2?

1 Answer

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