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2730 questions · page 88 of 137 · choose a topic above to narrow them down
SA-9709-P3-TR-004 4 marks L2 Standard Calculator Compound angles
Angles A A A B B B sin A = 3 5 \sin A=\tfrac35 sin A = 5 3 cos B = 12 13 \cos B=\tfrac{12}{13} cos B = 13 12 sin ( A + B ) \sin(A+B) sin ( A + B ) cos ( A − B ) \cos(A-B) cos ( A − B )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-TR-005 2 marks L2 Standard Calculator Compound angles
Given that tan A = 2 \tan A=2 tan A = 2 tan B = 1 3 \tan B=\tfrac13 tan B = 3 1 tan ( A + B ) \tan(A+B) tan ( A + B )
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-TR-009 3 marks L2 Standard Calculator R-form
Express 3 sin θ + 4 cos θ 3\sin\theta+4\cos\theta 3 sin θ + 4 cos θ R sin ( θ + α ) R\sin(\theta+\alpha) R sin ( θ + α ) R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘ α \alpha α
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-002 3 marks L2 Standard Calculator Magnitude
Find the unit vector in the direction of v = 2 i − 6 j + 3 k \mathbf{v}=2\mathbf{i}-6\mathbf{j}+3\mathbf{k} v = 2 i − 6 j + 3 k
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE-005 2 marks L2 Standard Calculator Perpendicular
The vectors ( 2 p − 1 ) \begin{pmatrix}2\\p\\-1\end{pmatrix} 2 p − 1 ( 3 1 4 ) \begin{pmatrix}3\\1\\4\end{pmatrix} 3 1 4 p p p
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SA-9709-P3-VE-006 3 marks L2 Standard Calculator Line equation
The line l l l A ( 1 , − 2 , 3 ) A(1,-2,3) A ( 1 , − 2 , 3 ) B ( 4 , 0 , 1 ) B(4,0,1) B ( 4 , 0 , 1 ) r = O A → + λ A B → \mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB} r = O A + λ A B l l l λ = 2 \lambda=2 λ = 2
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SA-9709-P3-VE1-003 3 marks L2 Standard Calculator Position vectors
A A A ( 1 , 4 , − 2 ) (1,4,-2) ( 1 , 4 , − 2 ) B B B ( 7 , 1 , 4 ) (7,1,4) ( 7 , 1 , 4 ) P P P A B AB A B A P : P B = 1 : 2 AP:PB=1:2 A P : P B = 1 : 2 P P P
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SA-9709-P3-VE2-001 3 marks L2 Standard Calculator Scalar product
Find the angle between a = 3 i + 4 k \mathbf{a}=3\mathbf{i}+4\mathbf{k} a = 3 i + 4 k b = 5 j + 12 k \mathbf{b}=5\mathbf{j}+12\mathbf{k} b = 5 j + 12 k
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SA-9709-P3-VE3-001 3 marks L2 Standard Calculator Line equation
The line through A ( 1 , 2 , 3 ) A(1,2,3) A ( 1 , 2 , 3 ) B ( 4 , 0 , 5 ) B(4,0,5) B ( 4 , 0 , 5 ) r = O A → + λ A B → \mathbf{r}=\overrightarrow{OA}+\lambda\overrightarrow{AB} r = O A + λ A B x x x y y y z = 11 z=11 z = 11
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SA-9709-P3-VE3-002 2 marks L2 Standard Calculator Line equation
The point ( 7 , a , 7 ) (7,a,7) ( 7 , a , 7 ) r = ( 1 2 3 ) + λ ( 3 − 2 2 ) \mathbf{r}=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}3\\-2\\2\end{pmatrix} r = 1 2 3 + λ 3 − 2 2 a a a
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SA-9709-P3-VE4-002 3 marks L2 Standard Calculator Intersections
l 1 : r = ( 1 1 0 ) + λ ( 1 0 1 ) l_1:\ \mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\1\end{pmatrix} l 1 : r = 1 1 0 + λ 1 0 1 l 2 : r = ( 2 0 1 ) + μ ( 0 1 1 ) l_2:\ \mathbf{r}=\begin{pmatrix}2\\0\\1\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix} l 2 : r = 2 0 1 + μ 0 1 1
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE4-G01 3 marks L2 Standard Calculator Intersections
The lines r = ( 1 1 1 ) + λ ( 1 2 0 ) \mathbf{r}=\begin{pmatrix}1\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix} r = 1 1 1 + λ 1 2 0 r = ( 0 5 − 1 ) + μ ( 1 − 1 1 ) \mathbf{r}=\begin{pmatrix}0\\5\\-1\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\1\end{pmatrix} r = 0 5 − 1 + μ 1 − 1 1 λ \lambda λ μ \mu μ
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-VE5-001 4 marks L2 Standard Calculator Foot of perpendicular
Find the coordinates of the foot of the perpendicular from A ( 5 , 1 , 0 ) A(5,1,0) A ( 5 , 1 , 0 ) r = ( 1 2 − 1 ) + λ ( 2 1 2 ) \mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\2\end{pmatrix} r = 1 2 − 1 + λ 2 1 2
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SA-9709-P3-VE5-002 2 marks L2 Standard Calculator Distances
The foot of the perpendicular from A ( 5 , 1 , 0 ) A(5,1,0) A ( 5 , 1 , 0 ) F ( 3 , 3 , 1 ) F(3,3,1) F ( 3 , 3 , 1 ) A A A
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SA-9709-P3-VE5-G01 3 marks L2 Standard Calculator Foot of perpendicular
The line l l l r = ( 1 2 − 1 ) + λ ( 2 1 2 ) \mathbf{r}=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\2\end{pmatrix} r = 1 2 − 1 + λ 2 1 2 A A A ( 5 , 1 , 0 ) (5,1,0) ( 5 , 1 , 0 ) P P P l l l λ \lambda λ λ \lambda λ A P → \overrightarrow{AP} A P l l l
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P4-EW3-001 4 marks L2 Standard Calculator Work–energy
A block of mass 5 5 5 10 10 10 6 6 6 15 15 15
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SA-9709-P4-EW3-003 4 marks L2 Standard Calculator Finding resistances
A ball of mass 0.5 0.5 0.5 20 20 20 18 m s − 1 18\ \text{m s}^{-1} 18 m s − 1
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SA-9709-P4-EW3-004 4 marks L2 Standard Calculator Finding resistances
A particle of mass 3 3 3 10 m s − 1 10\ \text{m s}^{-1} 10 m s − 1 12.5 12.5 12.5
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SA-9709-P4-FE-002 4 marks L2 Standard Calculator Equilibrium
A particle of weight 20 20 20 α \alpha α P P P 25 25 25 α \alpha α P P P
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SA-9709-P4-FE-004 2 marks L2 Standard Calculator Friction
A box of mass 5 5 5 μ = 0.4 \mu=0.4 μ = 0.4 15 15 15
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