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Trigonometry

Question 5 of 9

Area of Triangle

Question

In an isosceles triangle two sides are of length \(10\,\text{cm}\) and the area is \(15\,\text{cm}^2\). Find the size of the included angle. (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 4 steps
Triangle ABC, drawn to scale, with vertex A at the top left, B at the top right and C at the lower left. The two given sides are c = 10 from A to B and a = 10 from B to C; the third side, b, from A to C, is unlabelled. The included angle between the two given sides, at vertex B, is marked with an arc — this is the angle the solution finds.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2}\,ac \sin B \]
  2. Substitute values based on formula
    \[ 15\,\text{cm}^2 = \frac{1}{2}(10\,\text{cm})(10\,\text{cm})\sin B \]
  3. Solve for \(\sin B\)

    Evaluate \(\dfrac{1}{2}(10\,\text{cm})(10\,\text{cm})\sin B\):

    \[ 15\,\text{cm}^2 = (50\,\text{cm}^2)\sin B \]

    Divide both sides by \(50\,\text{cm}^2\):

    \[ \frac{15}{50} = \sin B \]

    Divide 15 by 50:

    \[ \sin B = 0.3 \]
  4. Final answer

    Compute for \(\sin^{-1} 0.3\) to get \(\angle B\):

    \[ B = 17.46^\circ \]

    That is, the size of the included angle B is 17.46 degrees, to 2 decimal places.