In triangle ABC, if a=5, B=45^° and C=60^°, find t – Class 11 Mathematics | ConceptFirst

Exercise 10(a) · Q14

\triangle ABC

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Step 1. Since the angles of a triangle sum to $180^\circ$ : $A=180^\circ-45^\circ-60^\circ=75^\circ$ .

Step 2. With one side ( $a=5$ ) and all three angles known, use

\Delta=\frac{a^2\sin B\sin C}{2\sin A}.

Step 3. Substitute $\sin45^\circ=\frac{\sqrt2}2,\ \sin60^\circ=\frac{\sqrt3}2,\ \sin75^\circ=\frac{\sqrt6+\sqrt2}4$ :

\Delta=\frac{25\left(\frac{\sqrt2}2\right)\left(\frac{\sqrt3}2\right)}{2\left(\frac{\sqrt6+\sqrt2}4\right)}=\frac{\frac{25\sqrt6}{4}}{\frac{\sqrt6+\sqrt2}{2}}=\frac{25\sqrt6}{4}\times\frac{2}{\sqrt6+\sqrt2}=\frac{25\sqrt6}{2(\sqrt6+\sqrt2)}.

Step 4. Rationalise by multiplying numerator and denominator by $(\sqrt6-\sqrt2)$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2018Set 1A7 marks
$P_1, P_2, P_3$
(i) $\dfrac{1}{P_1} + \dfrac{1}{P_2} + \dfrac{1}{P_3} = \dfrac{1}{r}$
(ii) $P_1 P_2 P_3 = \dfrac{(abc)^3}{8R^3} = \dfrac{8\Delta^3}{abc}$ .
›Reveal solutionSolution

Using Δ=½aP1=½bP2=½cP3, the altitudes are P1=2Δ/a etc.; combining with s=Δ/r and Δ=abc/4R gives both parts.

Since the area of the triangle can be computed using any side as base with its corresponding altitude:

$\Delta = \dfrac12 a P_1 = \dfrac12 b P_2 = \dfrac12 c P_3$

So: $P_1=\dfrac{2\Delta}{a},\quad P_2=\dfrac{2\Delta}{b},\quad P_3=\dfrac{2\Delta}{c}$

(i) Prove $\frac1{P_1}+\frac1{P_2}+\frac1{P_3}=\frac1r$ :

$\dfrac{1}{P_1}+\dfrac{1}{P_2}+\dfrac{1}{P_3} = \dfrac{a}{2\Delta}+\dfrac{b}{2\Delta}+\dfrac{c}{2\Delta} = \dfrac{a+b+c}{2\Delta} = \dfrac{2s}{2\Delta}=\dfrac{s}{\Delta}$

(where $s=\frac{a+b+c}2$ is the semi-perimeter). Using the standard inradius formula $r=\dfrac{\Delta}{s}$ , i.e. $\dfrac1r=\dfrac s\Delta$ :

$\dfrac{1}{P_1}+\dfrac{1}{P_2}+\dfrac{1}{P_3} = \dfrac{s}{\Delta} = \dfrac{1}{r}$

(ii) Prove $P_1P_2P_3=\dfrac{8\Delta^3}{abc}$ , and relate it to R:

$P_1P_2P_3 = \dfrac{2\Delta}{a}\cdot\dfrac{2\Delta}{b}\cdot\dfrac{2\Delta}{c} = \dfrac{8\Delta^3}{abc}$ — this part follows directly.

Now use the circumradius formula $\Delta=\dfrac{abc}{4R}$ , so $\Delta^3=\dfrac{(abc)^3}{64R^3}$ . Substituting:

…

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