(iii) (b) Find the area of $\triangle OPB$ .

Q: 37. (iii) (b) Find the area of $\triangle OPB$.

Area of $\triangle OPB = \frac{1}{2} \times OB \times PB$, since $\triangle OPB$ is right-angled at B (vertical $PB \perp$ horizontal $OB$). Substituting the horizontal distance OB and height PB obtained in the earlier parts of Q37 gives the required area in $\text{cm}^2$ (or $\text{m}^2$). (Exact figure not stated here as OB, PB values are not present in the supplied OCR — method is complete and exact.)

Some Applications of Trigonometry

Heights and distances - area of triangle

Official Answer

Area of $\triangle OPB = \frac{1}{2} \times OB \times PB$ , since $\triangle OPB$ is right-angled at B (vertical $PB \perp$ horizontal $OB$ ). Substituting the horizontal distance OB and height PB obtained in the earlier parts of Q37 gives the required area in $\text{cm}^2$ (or $\text{m}^2$ ). (Exact figure not stated here as OB, PB values are not present in the supplied OCR — method is complete and exact.)

area of triangleright-angled triangle1/2 base heightheights and distancesOB horizontal distancePB heighttrigonometry application

Marking Scheme

11 mark: correct identification that $\triangle OPB$ is right-angled and correct area formula $\text{Area} = \frac{1}{2} \times OB \times PB$ .
21 mark: correct substitution of the two lengths from parts (i)/(ii) and correct final value with proper square unit.
3Acceptable alternative: $\text{Area} = \frac{1}{2} \times OP \times PB$ if the right angle is at P, or Heron's formula if all three sides are known — full marks for any valid correct method.

Hint

$\triangle OPB$ is right-angled — area is simply $\frac{1}{2} \times (\text{horizontal distance}) \times (\text{vertical height})$ using the two lengths already found in the previous parts.

Quick Oral Answer

Since $\triangle OPB$ is right-angled at B, its area is just half the product of the two perpendicular sides — the horizontal distance OB and the height PB found in the earlier parts.

Analysis & Explanation

This final part of the trigonometry case study rewards recognising that the two lengths found earlier are simply the base and height of a right triangle.

Concept

$\triangle OPB$ is right-angled at B, since the vertical height PB always meets the horizontal ground OB at 90°.
Once OB and PB are known (from the earlier tanθ steps), $\text{Area} = \frac{1}{2} \times OB \times PB$ — no further trigonometry is needed.

Common mistakes

Dropping the 1/2 factor.
Using the slant line of sight OP (the hypotenuse) as the height instead of the true perpendicular side PB.
Writing the area without a squared unit.

Real-world

Surveyors and civil engineers use exactly this method to find the cross-sectional area of ramps, embankments and triangular plots from a measured baseline and an angle of elevation.

Honesty note: the numeric values of OB and PB depend on the angles/distances in the Q37 figure, which are not present in the supplied OCR; the method below is exact and standard, and the final number should be verified against the original paper.

Common Mistakes

1Forgetting the factor $\frac{1}{2}$ and computing $\text{base} \times \text{height}$ instead of half of it.
2Mixing up which side is the base and which is the perpendicular height, or using the hypotenuse OP as the height.
3Omitting the square unit (writing cm instead of $\text{cm}^2$ ) in an area answer.

Interesting Facts

The 'half base times height' area rule works for ANY triangle, not just right triangles — for a right triangle the two legs conveniently serve as base and height.

Surveyors historically computed inaccessible land areas exactly this way: measure a baseline and an angle of elevation, then reduce the field to right triangles.

The area formula $\frac{1}{2}ab\sin C$ generalises this: when $C = 90°$ , $\sin 90° = 1$ , and it collapses to $\frac{1}{2} \times \text{base} \times \text{height}$ .

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why is $\triangle OPB$ right-angled?

In a heights-and-distances set-up, PB is the vertical object (pole/tower/height) and OB is the horizontal ground distance from the observer to the base B. A vertical line meets a horizontal line at 90°, so the angle at B is a right angle, making OPB a right triangle with legs OB and PB.

Which sides are the base and the height?

The two sides forming the right angle are the base and the height. Here OB (horizontal) is the base and PB (vertical) is the perpendicular height. The hypotenuse OP is never used as the height in the $\frac{1}{2} \times \text{base} \times \text{height}$ formula.

Can I use Heron's formula instead?

Yes, if you know all three sides OB, PB and OP you may use Heron's formula and you will get the same answer, but for a right triangle $\frac{1}{2} \times OB \times PB$ is far quicker and less error-prone.