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secondary 2 | Maths
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pls help how to do its due tmr :(
Date Posted: 3 years ago
Views: 294
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hope this helps
Date Posted:
3 years ago
thank u
The angle range implies that θ is acute. So, you can draw a right-angled triangle with one of the acute angles being θ.
Then since sin θ = 4/5, and sin → opposite/hypotenuse, we can say the ratio of the length of the opposite to the hypotenuse is 4 : 5.
Since we have a right angled triangle, let the opposite be 4 units and the hypotenuse be 5 units.
Using Pythagoras Theorem,
(Opposite)² + (adjacent)² = (hypotenuse)²
4² + (adjacent)² = 5²
(adjacent)² = 5² - 4² = 9 = 3²
adjacent = 3 (since length is positive)
This is actually one of the most well known Pythagorean triples (3² + 4² = 5²)
cos θ = adjacent/hypotenuse = 3/5
tan θ = opposite/adjacent = 4/3
(Recall the mnemonic TOA CAH SOH)
Then since sin θ = 4/5, and sin → opposite/hypotenuse, we can say the ratio of the length of the opposite to the hypotenuse is 4 : 5.
Since we have a right angled triangle, let the opposite be 4 units and the hypotenuse be 5 units.
Using Pythagoras Theorem,
(Opposite)² + (adjacent)² = (hypotenuse)²
4² + (adjacent)² = 5²
(adjacent)² = 5² - 4² = 9 = 3²
adjacent = 3 (since length is positive)
This is actually one of the most well known Pythagorean triples (3² + 4² = 5²)
cos θ = adjacent/hypotenuse = 3/5
tan θ = opposite/adjacent = 4/3
(Recall the mnemonic TOA CAH SOH)
thank u
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