The correct answer is: $ \displaystyle 3\sqrt{3}$ cm Explanation: Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. Join OA and OP. Also, OP is a bisector line of ∠APC. ∴ ​∠APO = ∠CPO = 30°  Also,  OA⊥AP​ Tangent at any […]

$ \displaystyle \dfrac{120o}{2}$

The correct answer is: $ \displaystyle \text{84 c}{{\text{m}}^{2}}$ Explanation: The semi-perimeter $ \displaystyle s=\dfrac{{13+14+15}}{2}=21\text{ cm}$ Using Heron’s formula: Area $ \displaystyle \begin{array}{l}=\sqrt{{s\left( {s-a} \right)\left( {s-b} \right)\left( {s-c} \right)}}=\sqrt{{21\left( {21-13} \right)\left( {21-14} \right)\left( {21-15} \right)}}\\=\sqrt{{21\times 8\times 7\times 6}}=\sqrt{{7056}}=84\text{ c}{{\text{m}}^{2}}\end{array}$

4 kW Explanation: Total power output = Area × Solar irradiance × Efficiency = 25 m² × 800 W/m² × 0.20 = 4,000 W = 4 kW

$latex \displaystyle \begin{array}{l}(\text{Take g}=9.8m/{{s}^{2}})\\{{v}_{y}}=v\sin \theta \\so,80\sin {{60}^{\circ }}=80\times 0.866=69.28m/s\\\text{Maximum height }h=/2g={{(69.28)}^{2}}/2\times 9.8\approx 244.9m\end{array}$

$\displaystyle \begin{array}{l}\text{Resultant velocity i}\text{.e}\text{.}\\{{v}_{{re}}}=\sqrt{{{{v}_{s}}^{2}+{{v}_{r}}^{2}}}\\\text{So, }\sqrt{{{{5}^{2}}+{{3}^{2}}}}=\sqrt{{25+9}}=\sqrt{{34}}\\\approx 5.83~m/s\end{array}$

$\displaystyle \begin{array}{l}\text{Initial velocity }(u)=0m/s\\\text{Final velocity }(v)=20m/s\\\text{Time }(t)=5s\\\text{Acceleration }(a)=(v-u)/t=(20-0)/5=4m/{{s}^{2}}\\\text{Distance }({{s}_{1}})\text{ during acceleration}:\\{{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}+=0\times 5+\frac{1}{2}\times 4\times {{5}^{2}}=50cm\\\text{Deceleration phase:}\\\text{Initial velocity }(u)=20m/s\\\text{Final velocity }(v)=0m/s\\\text{Time }(t)=10s\\\text{Deceleration }(a)=(v-u)/t=(0-20)/10=-2m/{{s}^{2}}\\\text{Distance }({{s}_{2}})\text{during deceleration}:\\{{s}_{2}}=ut+\frac{1}{2}a{{t}^{2}}=20\times 10+\frac{1}{2}\times (-2)\times {{10}^{2}}=100m\\\end{array}$