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- Question11 and 12, Exercise 10.1
- ====== Question11 and 12, Exercise 10.1 ====== Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== If $\alpha$, $\beta$, $\gamma$ are the angle... align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-1-p11|Question 13 >]]</btn></text>
- Question 2, Exercise 10.2
- [4px,border:2px solid black]{\cos 2\theta=-\dfrac{119}{169}.}$$ =====Question 2(iii)===== If $\sin ... ^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}=\dfrac{119}{169}.\end{align} Now \begin{align}\tan 2\theta... \cos 2\theta }\\ &=\dfrac{\frac{-120}{169}}{\frac{119}{169}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{120}{119}}$$ <text align="left"><btn type="primary">[[fs
- Question 6, Exercise 10.2
- e half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfra... dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\
- Question, Exercise 10.1
- - \dfrac{5}{12}}{1+ \dfrac{20}{36}}=\dfrac{\dfrac{11}{12}}{ \dfrac{56}{36}}\\ \Rightarrow \quad \tan \
- Question 9 and 10, Exercise 10.1
- gn="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-1-p10|Question 11,12 >]]</btn></text>
- Question 13, Exercise 10.1
- lign="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-1-p10|< Question 11,12]]</btn></text>
- Question 2, Exercise 10.3
- {\circ }}}{2} \right)\\ &=-2\sin \left( \dfrac{{{118}^{\circ }}}{2} \right)\sin \left( \dfrac{-{{46}^