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The diagram shows two right-angled triangles, `A B C` and `A B D`,
where `A C=35 \ text{cm}`, `B D=93 \ text{cm}`, `/_ A C B=41^@` and ` /_ A D B=\theta`.
Calculate the size of angle `\theta`, to the nearest minute. (4 marks)
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`19^@6`′
`text{In}\ Delta ABC:`
`tan 41^@` | `=(AB)/35` | |
`AB` | `=35xxtan 41^@` | |
`=30.425…` |
`text{In}\ Delta ABD:`
`sin theta` | `=(AB)/(BD)` | |
`=(30.425…)/93` | ||
`:.theta` | `=sin^(-1)((30.425…)/93)` | |
`=19.09…` | ||
`=19^@6`′`\ \ text{(nearest minute)}` |
Two right-angled triangles, `ABC` and `ADC`, are shown.
Calculate the size of angle `theta`, correct to the nearest minute. (3 marks)
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`41°4′\ \ text{(nearest minute)}`
`text(Using Pythagoras in)\ DeltaACD:`♦♦♦ Mean mark 15%.
`AC^2` | `= 2.5^2 + 6^2` |
`= 42.25` | |
`:.AC` | `= 6.5\ text(cm)` |
`text(In)\ DeltaABC:`
`costheta` | `= 4.9/6.5` |
`theta` | `= cos^(−1)\ 4.9/6.5` |
`= 41.075…` | |
`= 41°4′31″` | |
`= 41°5′\ \ text{(nearest minute)}` |
The diagram shows a right-angled triangle.
What is the value of `theta`, to the nearest minute?
`B`
`tan theta` | `= text(opp)/text(adj)` |
`= 5.3/1.9` | |
`= 2.789…` |
`:. theta` | `= 70.277…^@` |
`=70°16^{′}39.8^{″}` | |
`= 70^@17^{′}` |
`=>B`
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i. `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`
`a^2 + b^2` | `= 5^2 + 12^2` |
`= 169` | |
`= 13^2` | |
`= c^2…\ text(as required.)` |
MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.
ii. `sin ∠ABC = 12/13`
`:.∠ABC` | `= 67.38…°` |
`=67°22^{′}48^{″}` | |
`= 67°23^{′}\ \ \ text{(nearest minute)}` |