E-Math - Trigonometry - 3 Dimension Problems - Open Door
a) the distance CE,
Since AD=BC=BE (Length of door)
As the triangle is NOT a Right Angle we will use Cosine Rule
\(\begin{array}{l}{c^2} = {a^2} + {b^2} - 2ab{\mathop{\rm Cos}\nolimits} C\\{x^2} = {1^2} + {1^2} - 2(1)(1){\mathop{\rm Cos}\nolimits} {38^o}\\{x^2} = 0.42398\\x = 0.65113 = 0.651m(3sig)\end{array}\)
b) the length AE,
Since it is a rectangular door. Angle ABE is 90 degrees.
We use Pythagoras Theorem.
\(\begin{array}{l}{c^2} = {a^2} + {b^2}\\{y^2} = {2^2} + {(1)^2}\\{c^2} = 5\\c = 2.2361 = 2.24m(3sig)\end{array}\)
c) \(\angle CAE\).
Use the values found in a) and b)
AE=AC=2.1033, EC=0.65113
Since the triangle is NOT a Right Angle,
we will use Cosine Rule.
\(\begin{array}{l}{c^2} = {a^2} + {b^2} - 2ab{\mathop{\rm Cos}\nolimits} C\\\\{0.65113^2} = {2.2361^2} + {2.2361^2} - 2(2.2361)(2.2361){\mathop{\rm Cos}\nolimits} C\\\\0.42397 = 10.000 - 10.000{\mathop{\rm Cos}\nolimits} C\\\\10.000{\mathop{\rm Cos}\nolimits} C = 10.000 - 0.42397\\\\CosC = \frac{{9.57602}}{{10.000}}\\\\C = {{\mathop{\rm Cos}\nolimits} ^{ - 1}}0.95760\\\\C = {16.74^o} = {16.7^o}\end{array}\)
When a Triangle is a Right Angle, use Sine, Cosine , Tangent and Pythagoras Theorem.
But when a Triangle is NOT a Right Angle, use Sine Rule and Cosine Rule (In the Formula List of the O level E-Math Exam Paper).
I suggest that you draw out the triangle to find the unknown side /angle as 3-D figure can be
visually confusing at times.
Additional Math (A-Maths) and Math (E-Math) Tutor in Woodlands, Chua Chu Kang,
Sembawang, Bukit Panjang, Yishun and Johor Bahru.
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