FBISE, NCP & NBF-aligned resources for Classes 9–12
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Math Chapter 6 : Test Yourself
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If $x=$ $\frac{1}{4} $, what is the value of $cos x$?
Reasoning: $cos x=sqrt{1-{sin}^{2} x} =$ $sqrt{1-frac{1}{16}} =sqrt{frac{15}{16}} =frac{sqrt{15}}{4}$
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$cos$ $\frac{2\pi}{3}= $?
Reasoning: $cos$ $frac{2pi}{3}=cos {120}^{o}=-frac{1}{2}$
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$\frac{1}{1 - sin \theta}+\frac{1}{1 + sin \theta}=$ ?
Reasoning: $frac{1}{1 - sin theta}+frac{1}{1 + sin theta}=frac{2}{1 - {sin}^{2} theta}=frac{2}{{cos}^{2} theta}=2 {sec}^{2} theta$
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$- 3-3 {tan}^{2} \theta=$ ?
Reasoning: $- 3-3 {tan}^{2} theta=- 3(1+{tan}^{2} theta)=- 3 {sec}^{2} theta$
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If $cos x=sin x$, then the value of $x $is:
Reasoning: $cos x=sin x Longrightarrow tan x=1 Longrightarrow x={45}^{o}$
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$4 {cosec}^{2} \theta-4 {cot}^{2} \theta-4 cos {0}^{o}=$ ?
Reasoning: ${cosec}^{2} theta-{cot}^{2} theta=1$So, $4 (1)-4 cos {0}^{o}=4-4=0$
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If $x=$ $\frac{a}{b} $, then $sin x=$ ?
Reasoning: If $x=$ $frac{a}{b} $, then:opposite $=a$, adjacent $=b$, hypotenuse $=sqrt{{a}^{2} + {b}^{2}}$So, $sin x=$ $frac{a}{sqrt{{a}^{2} + {b}^{2}}}$
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${50}^{o }{30}^{'}=$ ?
Reasoning: ${50}^{o }{30}^{'}=50+$ $frac{30}{60}={50.5}^{o}$
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$(1-{cos}^{2} \theta) {sec}^{2} \theta=$ ?
Reasoning: $(1-{cos}^{2} theta) {sec}^{2} theta={sin}^{2} theta$ . $frac{1}{{cos}^{2} theta}={tan}^{2} theta$
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In which quadrant do the angles between ${90}^{o}$ and ${180}^{o}$ lie?
Reasoning: Angles from ${90}^{o}$ to ${180}^{o}$ lie in the 2nd quadrant.
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The system in which the angles are measured in radians is called:
Reasoning: Radian system is called the circular system.
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${270}^{o}=$ ?
Reasoning: ${270}^{o}=$ $frac{270 pi}{180}=frac{3pi}{2}$
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What is the bearing of south-west direction line?
Reasoning: South-West direction $={180}^{o}+{45}^{o}={225}^{o}$
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If an automobile travels at the bearing of is$ {070}^{o}$ from city $A$ to city $B$, then the bearing from city $B$ to city $A$ is:
Reasoning: Reverse bearing $={070}^{o}+{180}^{o}={250}^{o}$
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A boat sails $4 km$ north from point $L$ to $M$, then it sails $3 km$ west from $M$ to $P$. What is the bearing of $P$ from $L$? Provide your answer to the nearest degree.
Reasoning: Displacement $=4 km$ north, $3 km$ westAngle west of north $={tan}^{-1} (3/4)approx{37}^{o}$Bearing $=360-37={323}^{o}$
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Reasoning: $cos x=\sqrt{1-{sin}^{2} x} =$ $\sqrt{1-\frac{1}{16}} =\sqrt{\frac{15}{16}} =\frac{\sqrt{15}}{4}$
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Reasoning: $cos$ $\frac{2\pi}{3}=cos {120}^{o}=-\frac{1}{2}$
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Reasoning: $\frac{1}{1 - sin \theta}+\frac{1}{1 + sin \theta}=\frac{2}{1 - {sin}^{2} \theta}=\frac{2}{{cos}^{2} \theta}=2 {sec}^{2} \theta$
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Reasoning: $- 3-3 {tan}^{2} \theta=- 3(1+{tan}^{2} \theta)=- 3 {sec}^{2} \theta$
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Reasoning: $cos x=sin x \Longrightarrow tan x=1 \Longrightarrow x={45}^{o}$
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Reasoning: ${cosec}^{2} \theta-{cot}^{2} \theta=1$So, $4 (1)-4 cos {0}^{o}=4-4=0$
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Reasoning: If $x=$ $\frac{a}{b} $, then:opposite $=a$, adjacent $=b$, hypotenuse $=\sqrt{{a}^{2} + {b}^{2}}$So, $sin x=$ $\frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$
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Reasoning: ${50}^{o }{30}^{'}=50+$ $\frac{30}{60}={50.5}^{o}$
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Reasoning: $(1-{cos}^{2} \theta) {sec}^{2} \theta={sin}^{2} \theta$ . $\frac{1}{{cos}^{2} \theta}={tan}^{2} \theta$
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Reasoning: ${270}^{o}=$ $\frac{270 \pi}{180}=\frac{3\pi}{2}$
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Reasoning: Displacement $=4 km$ north, $3 km$ westAngle west of north $={tan}^{-1} (3/4)\approx{37}^{o}$Bearing $=360-37={323}^{o}$
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