The area of acute-angled isosceles triangle PQR is 8cm squared. PQ = QR = 6cm. Find angle PQR to 2 decimal places. I tried it and got 4/9, but its not right. Please help. Thankyou
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You just didn't finish. $\displaystyle Area = \frac{1}{2}a^{2}sin{\theta}=8$ $\displaystyle 18sin{\theta}=8$ $\displaystyle sin{\theta}=\frac{4}{9}$ $\displaystyle {\theta}=sin^{-1}(\frac{4}{9})$
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