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Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
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extbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. ===== Question 1(i... a \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22... ha \cos \beta +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ }}\cos {{53}^... ha \cos \beta -\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{19}^{\circ }}\cos {{5}^{
Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
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minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\... minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\... minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\
Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
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ha$, $\beta$ and $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ... uad \beta $ and $\gamma $ are angles of triangle, therefore \begin{align}&\alpha +\beta +\gamma =180^\cir
Question 1, Exercise 10.2 @fsc-part1-kpk:sol:unit10
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extbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. =====Question1====
Question 4 and 5, Exercise 10.2 @fsc-part1-kpk:sol:unit10
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quadrant and $\sin$ is positive in 2nd quadrant, therefore \begin{align}\sin\dfrac{\theta }{2}&=-\sqrt{\
Question 1, Exercise 10.3 @fsc-part1-kpk:sol:unit10
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extbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. =====Question 1(i)
Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
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\end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\cos {{36}^{\circ }}-\cos {{82}^{\circ }}=2
Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
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\right)+\dfrac{1}{8}\sin {{50}^{\circ }},\quad \therefore \,\,\,\sin \left( 2(90)-\theta \right)=\sin
Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
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\right)+\dfrac{1}{8}\sin {{50}^{\circ }},\quad \therefore \,\,\,\sin \left( 2(90)-\theta \right)=\sin