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Question

In a quadrilateral ABCD, AB = 3.5 cm, BC = 8 cm, AD = 6 cm, angle ABD = 90° and angle CBD = 60°. Calculate

a) angle ADB

b) BD,

c) CD.

10 points goes to the person who can answer properly.

Thank you!

Answer 1

ADB = sin-1 (3.5/6) = 35.69

BD = sqrt(6^2 - 3.5^2) = 4.87

To find CD, use law of cosines CD^2 = 8^2 + 6.95^2 - 2(8)(6.95)cos(60)

Answer 2

Draw the quadrilateral with diagonal BD.

ABD is a triangle with a right angle at B.

sin(ADB) = 3.5/6 = 7/12

Angle ADB = arcsin(7/12) ≈ 35.7°

BD = √[(AD)² - (AB)²] = √(6² - 3.5²) = √23.75 ≈ 4.873

(CD)² = (BD)² + (BC)² - 2(BD)(BC)cos(CBD)

(CD)² = 23.75 + 8² - 2(√23.75)(8)cos(60°) ≈ 48.762823

CD ≈ 6.983

Answer 3

Hi,

Since ΔABD is a right triangle, then sin <ADB = 3.5/6 and <ADB = 35.7°.

Since  <ADB = 35.7°, cos 35.7° = BD/6. So BD = 6 cos 35.7° = 4.87.

To find CD, use the law of cosines:

c² = a² + b² - 2ab cos C

CD² = 8² + 4.87² - 2(8)(4.87) cos 60°

CD² = 48.7628

CD = 6.98

<ADB = 35.7°, BD = 4.87, CD = 6.98 <==ANSWERS

I hope that helps!! :-)